Factorise 8 x^{3}+27 y^{3}+125 z^{3}-90 x y z | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) The given expression is of the form $a^{3}+b^{3}+c^{3}-3 a b c$,where $a = 2x$,$b = 3y$,and $c = 5z$.Using the algebraic identity $a^{3}+b^{3}+c

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) The given expression is of the form $a^{3}+b^{3}+c^{3}-3 a b c$,where $a = 2x$,$b = 3y$,and $c = 5z$.
Using the algebraic identity $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$:
Substitute $a = 2x$,$b = 3y$,and $c = 5z$ into the identity:
$8 x^{3}+27 y^{3}+125 z^{3}-3(2x)(3y)(5z) = (2x+3y+5z)((2x)^{2}+(3y)^{2}+(5z)^{2}-(2x)(3y)-(3y)(5z)-(5z)(2x))$
$= (2x+3y+5z)(4x^{2}+9y^{2}+25z^{2}-6xy-15yz-10zx)$.

Factorise $8 x^{3}+27 y^{3}+125 z^{3}-90 x y z$.

Similar Questions

Factorise the following expression: $8x^3 + 343y^3 + 84x^2y + 294xy^2$.

With the help of the remainder theorem,find the remainder when the polynomial $p(x) = x^{3} + x^{2} - 26x + 24$ is divided by the divisor $x + 6$.

Factorise $: x^{3}-125$

By the Remainder Theorem,find the remainder when $p(x)$ is divided by $g(x)$,where $p(x) = x^{3} - 3x^{2} + 4x + 50$ and $g(x) = x - 3$.

The degree of the zero polynomial is

Vedclass Test Series

Exam Paper Generator

Online Exam Module