Cube 99 in Four Easy Numbers, Then Cancel What You Must Not
Learn to derive the identity for the cube of a sum, expand cubic brackets and cube numbers near 100, simplify a rational expression by factorising, and state the values that make it undefined.
How do you cube 99 without multiplying three times?
Working out the long way means two multiplications with carries all the way through. There is a better route, and it uses four numbers you can hold in your head.
is , and the cube of a difference has a fixed shape:
Four terms, alternating signs, no carries.
The plus version is just as quick. , and the digits of the answer even display the coefficients .
Those coefficients are the content of this page. gave in the first part of this chapter; the cube gives ; and the pattern continues upward into the binomial theorem of later classes.
This page covers the third part of the CBSE Class 9 Mathematics chapter on algebraic identities — deriving the cube identities, using them on brackets and on numbers, simplifying rational expressions by factorising, and saying exactly where such an expression has no value.
is , and the cube of a difference has a fixed shape:
Four terms, alternating signs, no carries.
The plus version is just as quick. , and the digits of the answer even display the coefficients .
Those coefficients are the content of this page. gave in the first part of this chapter; the cube gives ; and the pattern continues upward into the binomial theorem of later classes.
This page covers the third part of the CBSE Class 9 Mathematics chapter on algebraic identities — deriving the cube identities, using them on brackets and on numbers, simplifying rational expressions by factorising, and saying exactly where such an expression has no value.
Formula
How do you derive the identity for the cube of a sum?
Multiply the square identity by one more bracket. No memorising is needed if the derivation takes three lines.
Distribute the bracket across each term:
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Adding and collecting like terms, the and give , and the and give :
The difference version follows by replacing with . Odd powers of change sign and even powers do not:
Check both at , : , and . Then , and . Both correct.
The compact forms. Grouping the middle two terms gives a version that is often quicker to use:
Check the first at , : . Correct.
**The signs in alternate, unlike in .** The square ended on ; the cube ends on , because an odd power of a negative number stays negative. So the two cases behave differently, and the reason is the parity of the exponent — not a rule to be memorised separately but a consequence of .
**And is not .** With and : against . The gap of is exactly , which is the middle of the expansion doing its work.
Distribute the bracket across each term:
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Adding and collecting like terms, the and give , and the and give :
The difference version follows by replacing with . Odd powers of change sign and even powers do not:
Check both at , : , and . Then , and . Both correct.
The compact forms. Grouping the middle two terms gives a version that is often quicker to use:
Check the first at , : . Correct.
**The signs in alternate, unlike in .** The square ended on ; the cube ends on , because an odd power of a negative number stays negative. So the two cases behave differently, and the reason is the parity of the exponent — not a rule to be memorised separately but a consequence of .
**And is not .** With and : against . The gap of is exactly , which is the middle of the expansion doing its work.
How do you expand a cubic bracket and cube a number?
**Substitute into carefully, remembering that a coefficient inside the bracket gets cubed and squared along with the letter.
Worked example 1 — a coefficient on the first term.** Expand with , :
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Check at : the bracket gives , and the expansion gives . Correct.
Worked example 2 — with a minus and two letters. Expand with , :
Check at : , and . Correct.
**Worked example 3 — .** Take , :
**Worked example 4 — .** Now is subtracted, so the signs alternate:
Check the last digit: ends in , and ends in . Correct.
**Worked example 5 — .** With the powers of grow, so take care:
**Worked example 6 — from a base of .** With , :
**The term most often dropped is .** In worked example 5 it is , and leaving it out gives — a plausible-looking answer that is wrong by more than a thousand. Write all four terms down before adding any of them, which makes the omission impossible rather than unlikely.
Worked example 1 — a coefficient on the first term.** Expand with , :
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Check at : the bracket gives , and the expansion gives . Correct.
Worked example 2 — with a minus and two letters. Expand with , :
Check at : , and . Correct.
**Worked example 3 — .** Take , :
**Worked example 4 — .** Now is subtracted, so the signs alternate:
Check the last digit: ends in , and ends in . Correct.
**Worked example 5 — .** With the powers of grow, so take care:
**Worked example 6 — from a base of .** With , :
**The term most often dropped is .** In worked example 5 it is , and leaving it out gives — a plausible-looking answer that is wrong by more than a thousand. Write all four terms down before adding any of them, which makes the omission impossible rather than unlikely.
How do you simplify a rational algebraic expression?
Factorise the numerator and the denominator completely, then cancel the factors they share. Nothing can be cancelled until both parts are products.
Worked example 1 — a difference of squares over a quadratic. Simplify
The numerator is a difference of squares; the denominator splits with the pair and :
Check at : the original is , and the simplified form is . Correct.
Worked example 2 — a factor that cancels entirely. Simplify :
Worked example 3 — using a cube. Simplify . Since , the numerator is a sum of cubes and factorises as :
Check at : the original is , and the simplified form gives . Correct.
Worked example 4 — cancel after taking out a common factor. Simplify :
Only whole factors cancel, never individual terms. In the in the numerator cannot be cancelled against the in the denominator, because neither is a factor of its expression — both are terms inside a sum. Testing at a number settles it instantly: at the fraction is , while cancelling the would suggest . Different numbers, so the cancelling was illegal.
The rule in one sentence. Cancellation is division, and division distributes over a product, not over a sum. That is why the factorising has to come first and why an expression that will not factorise cannot be simplified at all.
Worked example 1 — a difference of squares over a quadratic. Simplify
The numerator is a difference of squares; the denominator splits with the pair and :
Check at : the original is , and the simplified form is . Correct.
Worked example 2 — a factor that cancels entirely. Simplify :
Worked example 3 — using a cube. Simplify . Since , the numerator is a sum of cubes and factorises as :
Check at : the original is , and the simplified form gives . Correct.
Worked example 4 — cancel after taking out a common factor. Simplify :
Only whole factors cancel, never individual terms. In the in the numerator cannot be cancelled against the in the denominator, because neither is a factor of its expression — both are terms inside a sum. Testing at a number settles it instantly: at the fraction is , while cancelling the would suggest . Different numbers, so the cancelling was illegal.
The rule in one sentence. Cancellation is division, and division distributes over a product, not over a sum. That is why the factorising has to come first and why an expression that will not factorise cannot be simplified at all.
Where does a rational expression have no value at all?
Wherever the original denominator is zero — and the values must be read off before any cancelling, because cancelling hides them.
Worked example 1. For , factorise the denominator:
This is zero when or , so the expression is **undefined at and **.
The simplified form was , which looks undefined only at . But at the original expression is , which has no value. The restriction survives the cancelling even though the factor does not, and the correct answer is
Worked example 2. simplifies to , which has a value everywhere. The original does not: at it is . So the simplification holds **for **, and at that one point the two expressions genuinely differ — one is undefined and the other equals .
Worked example 3. For , obtained above from , the original denominator is zero only at . So the restriction is .
Worked example 4 — a denominator with no zeros. For , the denominator is at least for every real , so the expression is defined everywhere. Not every rational expression carries a restriction.
How to verify a claimed identity. Expand both sides completely and compare term by term. Take the claim . Expanding the left gives , which differs from the right by — not zero in general, so the claim is false.
A single counter-example is enough to refute it: at , the left is and the right is .
But a single example is never enough to prove one. The claim happens to hold at , , and at , , and at , — infinitely many cases, all worthless as proof. An identity must be established by expanding both sides, and verification by substitution can only ever disprove. That asymmetry is worth remembering, because "I checked it with numbers and it worked" is the most common wrong justification in this chapter.
Worked example 1. For , factorise the denominator:
This is zero when or , so the expression is **undefined at and **.
The simplified form was , which looks undefined only at . But at the original expression is , which has no value. The restriction survives the cancelling even though the factor does not, and the correct answer is
Worked example 2. simplifies to , which has a value everywhere. The original does not: at it is . So the simplification holds **for **, and at that one point the two expressions genuinely differ — one is undefined and the other equals .
Worked example 3. For , obtained above from , the original denominator is zero only at . So the restriction is .
Worked example 4 — a denominator with no zeros. For , the denominator is at least for every real , so the expression is defined everywhere. Not every rational expression carries a restriction.
How to verify a claimed identity. Expand both sides completely and compare term by term. Take the claim . Expanding the left gives , which differs from the right by — not zero in general, so the claim is false.
A single counter-example is enough to refute it: at , the left is and the right is .
But a single example is never enough to prove one. The claim happens to hold at , , and at , , and at , — infinitely many cases, all worthless as proof. An identity must be established by expanding both sides, and verification by substitution can only ever disprove. That asymmetry is worth remembering, because "I checked it with numbers and it worked" is the most common wrong justification in this chapter.
Exam tip
Exam tip: write all four terms, and state the excluded values
Write the four terms of a cube before adding any of them: , , , . The dropped one is almost always .
Cube the coefficient too. In the first term is , not , and is .
**Signs alternate in ** — it ends on , unlike which ends on .
**Check every expansion by substituting .** at gives , and .
Use the compact form when it is faster: .
Verify the last digit of a numerical cube: must end in .
Factorise fully before cancelling. Nothing can be cancelled out of a sum — only out of a product.
Never cancel a term. does not reduce; at it is , not .
Read the excluded values off the ORIGINAL denominator, before cancelling. For they are and , even though the simplified form only shows the second.
Write the restriction beside the answer — marks are given for it and it is the easiest one in the question to forget.
And to refute a claimed identity, one counter-example suffices; to prove one, expand both sides — substitution can never prove an identity.
Cube the coefficient too. In the first term is , not , and is .
**Signs alternate in ** — it ends on , unlike which ends on .
**Check every expansion by substituting .** at gives , and .
Use the compact form when it is faster: .
Verify the last digit of a numerical cube: must end in .
Factorise fully before cancelling. Nothing can be cancelled out of a sum — only out of a product.
Never cancel a term. does not reduce; at it is , not .
Read the excluded values off the ORIGINAL denominator, before cancelling. For they are and , even though the simplified form only shows the second.
Write the restriction beside the answer — marks are given for it and it is the easiest one in the question to forget.
And to refute a claimed identity, one counter-example suffices; to prove one, expand both sides — substitution can never prove an identity.
Did you know
Why the coefficients 1, 3, 3, 1 show up in the answer's digits
Look at what the cube identity produces for :
Split the digits into pairs from the right: . The coefficients are sitting in the answer, spaced out by zeros.
The same happens one power down. , and the digits carry . And one power up:
with digits .
Those rows — then then then — are the rows of Pascal's triangle, where each entry is the sum of the two above it. Cubing displays a row because , so every term of the expansion is a coefficient multiplied by a power of , and powers of line the coefficients up in neat two-digit slots.
The display breaks as soon as a coefficient needs more than two digits. would want the row , and and cannot fit in a two-digit slot without carrying into the neighbour — so the digits stop being readable as coefficients.
It is also why does not show the pattern: carries , which is multiplied by — the powers of mixed in.
So the tidy digits of are a coincidence of base ten and the number , and the coefficients underneath them are not a coincidence at all. They are the same the derivation produced by collecting like terms.
Split the digits into pairs from the right: . The coefficients are sitting in the answer, spaced out by zeros.
The same happens one power down. , and the digits carry . And one power up:
with digits .
Those rows — then then then — are the rows of Pascal's triangle, where each entry is the sum of the two above it. Cubing displays a row because , so every term of the expansion is a coefficient multiplied by a power of , and powers of line the coefficients up in neat two-digit slots.
The display breaks as soon as a coefficient needs more than two digits. would want the row , and and cannot fit in a two-digit slot without carrying into the neighbour — so the digits stop being readable as coefficients.
It is also why does not show the pattern: carries , which is multiplied by — the powers of mixed in.
So the tidy digits of are a coincidence of base ten and the number , and the coefficients underneath them are not a coincidence at all. They are the same the derivation produced by collecting like terms.
Exam relevance
How do cube identities and rational expressions feed into JEE Main?
Because the cube expansion is the accessible case of a theorem examined directly, and the excluded-value habit is the whole of what "domain" means later.
This is the foundation for Class 11 Mathematics Binomial Theorem, examined in JEE Main. The coefficients found here by collecting like terms are the binomial coefficients , and the general theorem replaces the cube with any power . A student who has derived by multiplying by has seen exactly why the coefficients are sums of the previous row.
The cube identities themselves are reused in Class 10 Polynomials and Class 11 Complex Numbers. The related factorisations and — used in worked example 3 of the simplification section — appear constantly, and the symmetric identity is a standard JEE Main tool for expressions where .
The excluded values are the domain. Class 11 Relations and Functions asks for the domain of a function, and for a rational function that means precisely the question asked here: where is the denominator zero. The question type is standard and the working is identical — factorise the denominator, set each factor to zero.
And the cancelled factor becomes a limit. Class 11 Limits and Derivatives is built on expressions like , which is undefined at and yet approaches as approaches . The Class 9 observation that the simplified form and the original differ at exactly one point is the entire idea of a removable discontinuity, met properly in Class 12 Continuity and Differentiability. Getting that point right now makes that chapter a restatement rather than a surprise.
What the questions look like. For board work, expect expand a cubic bracket, evaluate a numerical cube, simplify a rational expression, state where it is undefined, and verify whether a given statement is an identity — the last one asked as a short proof. For JEE Main, the cube identities appear inside simplification steps, the factorisations of appear in limits and in algebraic manipulation, and the domain question appears directly.
How board and competitive emphasis differ. A board paper rewards the written derivation and the stated restriction. A competitive paper assumes both and tests whether the right identity is recognised in one glance — usually hiding as or .
The single trap that costs the most marks. Reading the excluded values off the simplified expression. In , the value is excluded and leaves no trace in the answer. Questions are built on exactly that disappearance, and the only defence is to factorise the original denominator and write the restrictions down before simplifying anything.
This is the foundation for Class 11 Mathematics Binomial Theorem, examined in JEE Main. The coefficients found here by collecting like terms are the binomial coefficients , and the general theorem replaces the cube with any power . A student who has derived by multiplying by has seen exactly why the coefficients are sums of the previous row.
The cube identities themselves are reused in Class 10 Polynomials and Class 11 Complex Numbers. The related factorisations and — used in worked example 3 of the simplification section — appear constantly, and the symmetric identity is a standard JEE Main tool for expressions where .
The excluded values are the domain. Class 11 Relations and Functions asks for the domain of a function, and for a rational function that means precisely the question asked here: where is the denominator zero. The question type is standard and the working is identical — factorise the denominator, set each factor to zero.
And the cancelled factor becomes a limit. Class 11 Limits and Derivatives is built on expressions like , which is undefined at and yet approaches as approaches . The Class 9 observation that the simplified form and the original differ at exactly one point is the entire idea of a removable discontinuity, met properly in Class 12 Continuity and Differentiability. Getting that point right now makes that chapter a restatement rather than a surprise.
What the questions look like. For board work, expect expand a cubic bracket, evaluate a numerical cube, simplify a rational expression, state where it is undefined, and verify whether a given statement is an identity — the last one asked as a short proof. For JEE Main, the cube identities appear inside simplification steps, the factorisations of appear in limits and in algebraic manipulation, and the domain question appears directly.
How board and competitive emphasis differ. A board paper rewards the written derivation and the stated restriction. A competitive paper assumes both and tests whether the right identity is recognised in one glance — usually hiding as or .
The single trap that costs the most marks. Reading the excluded values off the simplified expression. In , the value is excluded and leaves no trace in the answer. Questions are built on exactly that disappearance, and the only defence is to factorise the original denominator and write the restrictions down before simplifying anything.
Key takeaways
Cube identities and rational expressions: quick revision
- Derive, don't memorise: , and collecting like terms gives .
- **** — replace by ; odd powers change sign.
- Signs alternate in the cube, ending on ; the square ended on .
- Compact forms: and .
- ** is not **: at , it is against , and the gap is .
- , checked at as .
- , checked at as .
- Numerical cubes: ; ; ; .
- **The term most often dropped is — write all four before adding.
- To simplify a rational expression, factorise numerator and denominator fully, then cancel shared factors**.
- .
- ; using .
- .
- Never cancel a term. does not reduce — at it is , not .
- Read excluded values off the ORIGINAL denominator. is undefined at and , though the simplified form hides the first.
- holds **for **; at that point the original is .
- has no excluded values, since always.
- To refute a claimed identity, one counter-example is enough; to prove one, expand both sides. Substitution can never prove an identity.
- The digits of , and display the rows and and of Pascal's triangle.
Cube a two-digit number of your choice from a round base, then check it against ordinary multiplication — and notice which of the four terms you were tempted to skip.
- **** — replace by ; odd powers change sign.
- Signs alternate in the cube, ending on ; the square ended on .
- Compact forms: and .
- ** is not **: at , it is against , and the gap is .
- , checked at as .
- , checked at as .
- Numerical cubes: ; ; ; .
- **The term most often dropped is — write all four before adding.
- To simplify a rational expression, factorise numerator and denominator fully, then cancel shared factors**.
- .
- ; using .
- .
- Never cancel a term. does not reduce — at it is , not .
- Read excluded values off the ORIGINAL denominator. is undefined at and , though the simplified form hides the first.
- holds **for **; at that point the original is .
- has no excluded values, since always.
- To refute a claimed identity, one counter-example is enough; to prove one, expand both sides. Substitution can never prove an identity.
- The digits of , and display the rows and and of Pascal's triangle.
Cube a two-digit number of your choice from a round base, then check it against ordinary multiplication — and notice which of the four terms you were tempted to skip.