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Factorise each of the following:
(i) $ 27 y^{3}+125 z^{3} $
(ii) $ 64 m^{3}-343 n^{3} $
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 $64m^3-343n^3= (4m-7n)(16m^2+28mn+49n^2)$.
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