Factorise $8x^{3} + 125y^{3} + 343 - 210xy$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The given expression is of the form $a^{3} + b^{3} + c^{3} - 3abc$,where $a = 2x$,$b = 5y$,and $c = 7$.
We know the identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Here,$3abc = 3(2x)(5y)(7) = 210xy$.
Substituting the values into the identity:
$8x^{3} + 125y^{3} + 343 - 210xy = (2x + 5y + 7)((2x)^{2} + (5y)^{2} + (7)^{2} - (2x)(5y) - (5y)(7) - (7)(2x))$.
Simplifying the terms:
$= (2x + 5y + 7)(4x^{2} + 25y^{2} + 49 - 10xy - 35y - 14x)$.

Explore More

Similar Questions

Factorise the following quadratic polynomial by splitting the middle term:
$x^{2}-4x-77$

Divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$ and find the quotient and the remainder.

$4x^{2} - 20x + 25 = (\ldots \ldots \ldots)^{2}$

If $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ is divided by $(x + 1)$,the remainder is $19$. Find the value of $a$. Also,find the remainder when $p(x)$ is divided by $(x + 2)$.

Difficult
View Solution

If $x^{2}-8x-20=(x+a)(x+b),$ then $ab=\ldots \ldots \ldots$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo