cubic-identities-and-rational-expressions MCQs for UPSC Prelims

28 practice questions on cubic-identities-and-rational-expressions from the Exploring Algebraic Identities section of the UPSC Prelims syllabus. 28 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

9 Easy 14 Medium 5 Hard

Sample questions

Q1
easy
Simplify $\frac{x^2-9}{x+3}$, where $x\neq -3$.
  1. A $x-3$
  2. B $x+3$
  3. C $x^2-3$
  4. D $\frac{x-3}{x+3}$
Show answer and explanation

Correct answer: A - $x-3$

The numerator is a difference of squares, so $x^2-9=(x+3)(x-3)$ and the common factor $x+3$ cancels, leaving $x-3$. At $x=2$ the original is $\frac{4-9}{5}=-1$, and $2-3=-1$. The other answers give $5$, $1$ and $-\frac{1}{5}$ at $x=2$, because they cancel the wrong factor or cancel only part of it.
Q2
medium
Simplify $\frac{x^2-7x+12}{5x^2+5x-100}$, assuming no denominator is zero.
  1. A $\frac{x-3}{5(x+5)}$
  2. B $\frac{x-4}{5(x+5)}$
  3. C $\frac{x-3}{x+5}$
  4. D $\frac{x-3}{5(x-5)}$
Show answer and explanation

Correct answer: A - $\frac{x-3}{5(x+5)}$

The numerator factorises as $(x-3)(x-4)$, and the denominator is $5(x^2+x-20)=5(x+5)(x-4)$, so the common factor $x-4$ cancels. At $x=0$ the original is $\frac{12}{-100}=-0.12$, and $\frac{-3}{25}=-0.12$. The other answers give $0.12$, $-0.16$ and $-0.6$ at $x=0$, because they cancel the wrong bracket or drop the factor $5$.
Q3
medium
Statement I: the expansion of $(a-b)^3$ can be obtained from that of $(a+b)^3$ by replacing $b$ with $-b$. Statement II: in the expansion of $(a-b)^3$, the terms $a^3$ and $3ab^2$ both carry a plus sign. Which is correct?
  1. A Only Statement II is true
  2. B Neither statement is true
  3. C Both statements are true
  4. D Only Statement I is true
Show answer and explanation

Correct answer: C - Both statements are true

Replacing $b$ by $-b$ in $a^3+3a^2b+3ab^2+b^3$ gives $a^3-3a^2b+3ab^2-b^3$, since $(-b)^2=b^2$ while $-b$ and $(-b)^3$ are negative; that is exactly $(a-b)^3$, so the first statement holds. The same expansion shows $a^3$ and $3ab^2$ keeping plus signs, so the second holds as well. A check at $a=3$, $b=1$ gives $27-27+9-1=8=(3-1)^3$.
Q4
easy
Which factorisation of $x^3+y^3+z^3-3xyz$ is correct?
  1. A $(x+y+z)(x^2+y^2+z^2+xy+yz+zx)$
  2. B $(x+y+z)(x^2+y^2+z^2)$
  3. C $(x-y-z)(x^2+y^2+z^2-xy-yz-zx)$
  4. D $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Show answer and explanation

Correct answer: D - $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Taking $x=1$, $y=2$, $z=3$, the expression is $1+8+27-18=18$. The correct factorisation gives $6\times(14-2-6-3)=6\times 3=18$. Making the three products positive gives $6\times 25=150$, leaving them out gives $6\times 14=84$, and starting the first bracket with $x-y-z$ gives $-4\times 3=-12$.
Q5
hard
Simplify $\frac{a^3-b^3}{a^2-b^2}$, where $a$ and $b$ are unequal positive numbers.
  1. A $\frac{a+b}{a^2+ab+b^2}$
  2. B $\frac{a^2-ab+b^2}{a+b}$
  3. C $\frac{a^2+ab+b^2}{a+b}$
  4. D $a-b$
Show answer and explanation

Correct answer: C - $\frac{a^2+ab+b^2}{a+b}$

Factorising gives $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$, and the common factor $a-b$ cancels to leave $\frac{a^2+ab+b^2}{a+b}$. At $a=3$, $b=1$ the original is $\frac{26}{8}=3.25$, and $\frac{13}{4}=3.25$. Using $-ab$ gives $1.75$, cancelling both brackets wrongly gives $2$, and turning the fraction upside down gives about $0.31$.
Q6
medium
Expand $(3x-2)^3$.
  1. A $27x^3-54x^2+36x-8$
  2. B $27x^3+54x^2-36x-8$
  3. C $27x^3-18x^2+12x-8$
  4. D $27x^3-54x^2+36x+8$
Show answer and explanation

Correct answer: A - $27x^3-54x^2+36x-8$

With $a=3x$ and $b=2$, the terms are $27x^3$, $-3(9x^2)(2)=-54x^2$, $+3(3x)(4)=36x$ and $-8$. At $x=1$ the cube is $1^3=1$, and $27-54+36-8=1$. Ending in $+8$ gives $17$, forgetting the factor $3$ in the middle terms gives $13$, and reversing the middle signs gives $37$.
Q7
medium
Assertion (A): if $x+y+z=0$ then $x^3+y^3+z^3=0$. Reason (R): $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: D - A is false, but R is true

(R) is the standard identity and is true. Putting $x+y+z=0$ into it makes the right side zero, which gives $x^3+y^3+z^3=3xyz$, not $0$. With $x=1$, $y=2$ and $z=-3$ the sum is zero, yet $1+8-27=-18$, which is exactly $3(1)(2)(-3)$. So the reason holds while the assertion does not.
Q8
medium
To derive the expansion of $(a+b)^3$, we write it as $(a+b)(a^2+2ab+b^2)$. Multiplying $a$ by the bracket gives $a^3+2a^2b+ab^2$. What does multiplying $b$ by the bracket give?
  1. A $a^2b+ab^2+b^3$
  2. B $a^2b+2ab^2$
  3. C $ab^2+2ab+b^3$
  4. D $a^2b+2ab^2+b^3$
Show answer and explanation

Correct answer: D - $a^2b+2ab^2+b^3$

Multiplying each term of $a^2+2ab+b^2$ by $b$ gives $a^2b$, $2ab^2$ and $b^3$. Adding this to $a^3+2a^2b+ab^2$ produces $a^3+3a^2b+3ab^2+b^3$, the correct cube. At $a=3$, $b=2$ this part equals $18+24+8=50$, and the first part is $75$, making $125=(3+2)^3$. The other forms give $38$, $32$ and $42$ at those values.

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