Factorise the following: 64 a^{3}-27 b^{3}-14 | vedclass

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

Question

(A) The given expression is $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$.This can be rewritten as $(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$

Vedclass · Accepted Answer

$(4a - 3b)^3$

Vedclass · Answer

(A) The given expression is $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$.This can be rewritten as $(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$.This expression follows the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3x^{2}y+3xy^{2}$,where $x = 4a$ and $y = 3b$.Therefore,the expression simplifies to $(4 a-3 b)^{3}$.Thus,the factors are $(4 a-3 b)(4 a-3 b)(4 a-3 b)$.

Factorise the following: $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$

Similar Questions

Find the remainder when $x^{3}+3x^{2}+3x+1$ is divided by $5+2x$.

Factorise $6x^2 + 17x + 5$ by splitting the middle term,and by using the Factor Theorem.

Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: $p(x) = 2x^3 + x^2 - 2x - 1$,$g(x) = x + 1$.

Factorise the following using appropriate identities: $9x^{2}+6xy+y^{2}$

Factorise each of the following: $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

Vedclass Test Series

Exam Paper Generator

Online Exam Module