# Factorise each of the following:(i) $27 y^{3}+125 z^{3}$(ii) $64 m^{3}-343 n^{3}$

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To do:

We have to factorise each of the given expressions.

Solution:

We know that,

$a^3 + b^3=(a+b)^3-3ab(a+b)$

$=(a+b)[(a+b)^2-3ab]$

$=(a+b)[a^2+b^2+2ab-3ab]$

$=(a+b)(a^2+b^2-ab)$............(i)

$a^3-b^3=(a-b)^3+3ab(a-b)$

$=(a-b)[(a-b)^2+3ab]$

$=(a-b)[a^2+b^2-2ab+3ab]$

$=(a-b)(a^2+b^2+ab)$............(ii)

Therefore,

(i) $27y^3+125z^3$ can be written as,

$27y^3+125z^3=(3y)^3+(5z)^3$

This implies,

$(3y)^3+(5z)^3= (3y+5z)[(3y)^2-(3y)(5z)+(5z)^2]$         [Using (i)]

$= (3y+5z)(9y^2-15yz+25z^2)$

Hence $27y^3+125z^3=(3y+5z)(9y^2-15yz+25z^2)$.

(ii) $64m^3-343n^3$ can be written as,

$64m^3-343n^3 = (4m)^3-(7n)^3$

$= (4m-7n)[(4m)^2+(4m)(7n)+(7n)^2]$          [Using (ii)]

$= (4m-7n)(16m^2+28mn+49n^2)$

Hence $64m^3-343n^3= (4m-7n)(16m^2+28mn+49n^2)$.

Updated on 10-Oct-2022 13:39:07