Download App

Polynomials — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsPolynomials1 Mark
MCQ
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}=$
  • A
    3(a + b)( b+ c)(c + a)
  • B
    3(a - b)(b - c)(c - a)
  • C
    (a - b)(b - c)(c - a)
  • D
    None of these.
✓

Answer

  1. None of these.
    Solution:
    If a + b + c = 0 then, a3 + b3 + c3 = 3abc
    Now, (a2 - b2) + (b2 - c2) + (c2 - a2) = a2 - b2 + b2 - c2 + c2 - a2 = 0
    ⇒ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
    Again, (a - b) + (b - c) + (c - a) = a - b + b - c + c - a = 0
    ⇒ (a - b)3 + (b - c)3 + (c - a)3 = 3(a - b)(b - c)(c - a)
    Thus, we have
    $\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
    $=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
    $=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
    $=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
    Hence, correct option is (d).

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q" from Polynomials→Polynomials — all question types→Maths STD 9 question papers→Gujarat Board STD 9 question papers→Gujarat Board question paper generator→

Similar questions

In $\triangle\text{ABC},$ if D is the midpoint of BC and E is the midpoint of AD, then $\text{ar}(\triangle\text{BED})=?$

  1. $\frac{1}{2}\text{ar}(\triangle\text{ABC})$

  2. $\frac{1}{3}\text{ar}(\triangle\text{ABC})$

  3. $\frac{1}{4}\text{ar}(\triangle\text{ABC})$

  4. $\frac{2}{3}\text{ar}(\triangle\text{ABC})$

→
The product (a + b)(a - b)(a2 - ab + b2)(a2 + ab + b2) is equal to:
→
The side BC of $\triangle\text{ABC}$ is produced to a point D. The bisector of $\angle\text{A}$ meets side in L. If $\angle\text{ABC} = 30^\circ$ and $\angle\text{ACD} = 115^\circ$ then $\angle\text{ALC} =$
→
$\frac{1}{\big(3+2\sqrt{2}\big)}=?$
  1. $\frac{3-2\sqrt{2}}{17}$
  2. $\frac{\big(3-2\sqrt{2}\big)}{13}$
  3. $\big(3-2\sqrt{2}\big)$
  4. None of these.
→
In the given figure, AOB is a diameter of a circle and CD || AB. If $\angle\text{BAD}=30^\circ$ then $\angle\text{CAD}=?$
→
In Fig. $\angle B O D=$
Image
→
Write the correct answer in the following:
Euclid stated that all right angles are equal to each other in the form of:
  1. An axiom.
  2. A definition.
  3. A postulate.
  4. A proof.
→
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is
→
The length of the longest rod that can be placed in a room of dimension (10m × 10m × 5m) is:
→
If the length of a chord of a circle is 16cm and is at a distance of 15cm from the centre of the circle, then the radius of the circle is:
→