Factorise the following:
$1-64 a^{3}-12 a+48 a^{2}$
$1-64 a^{3}-12 a+48 a^{2}$
(A) The given expression is $1-64 a^{3}-12 a+48 a^{2}$.
We can rewrite this expression as $1^{3}-(4 a)^{3}-3(1)(4 a)(1-4 a)$.
This matches the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3xy(x-y)$,where $x=1$ and $y=4a$.
Therefore,the expression simplifies to $(1-4 a)^{3}$.
Thus,the factorised form is $(1-4 a)(1-4 a)(1-4 a)$.
We can rewrite this expression as $1^{3}-(4 a)^{3}-3(1)(4 a)(1-4 a)$.
This matches the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3xy(x-y)$,where $x=1$ and $y=4a$.
Therefore,the expression simplifies to $(1-4 a)^{3}$.
Thus,the factorised form is $(1-4 a)(1-4 a)(1-4 a)$.
Explore More
Similar Questions
The factorisation of $4 x^{2}+8 x+3$ is
Easy
View SolutionFrom the following polynomials,find out which of them has $(x+1)$ as a factor:
$P(x) = 6x^3 + 11x^2 - 5x - 12$
$P(x) = 6x^3 + 11x^2 - 5x - 12$
Medium
View SolutionFactorise the following quadratic polynomial by splitting the middle term:
$x^{2}-4x-77$
$x^{2}-4x-77$
Easy
View SolutionFactorise: $x^{3}+x^{2}-4x-4$
Medium
View SolutionDetermine whether $(x + 1)$ is a factor of the polynomial $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo