Without finding cubes, factorise (x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}

English

Gujarati

Hindi

2. Polynomials

hard

Without finding the cubes, factorise

$(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$

Option A

Option B

Option C

Option D

Solution

Let $x-2 y=a, 2 y-3 z=b$ and $3 z-c=x$

$\therefore \quad a+b+c=x-2 y+2 y-3 z+3 z-x=0$

$\Rightarrow \quad a^{3}+b^{3}+c^{3}=3 a b c$

Hence, $(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$

$=3(x-2 y)(2 y-3 z)(3 z-x)$

Standard 9

Mathematics

Share

Similar Questions

Write the following cubes in expanded form

$(4 x-3 y)^{3}$

easy

Determine which of the following polynomials has $x-2$ a factor :

$3 x^{2}+6 x-24$

$4 x^{2}+ x-2$

medium

Factorise

$25 x^{2}+9 y^{2}+64+30 x y-48 y-80 x$

easy

With the help of the remainder theorem, find the remainder when the polynomial $x^{3}+x^{2}-26 x+24$ is divided by each of the following divisors

$x-6$

easy

Factorise $: 121 x^{2}-289 y^{2}$

easy