Question

factorise: 27y cube+125z cube

Answer 1

Factorization of 27 $y^{3}+125 z^{3}\ \text{is}\ \bold{(3 y+5 z)\left(9 y^{2}+25 z^{2}-15 y z\right)}$

Given:  

$27 y^{3}+125 z^{3}$

To find:

Factorization of $27 y^{3}+125 z^{3}$

Solution:

We have to factorize -

$27 y^{3}+125 z^{3}=(3 y)^{3}+(5 z)^{3} \rightarrow(1)$ [ Since 27 = cube of 3, 125 = cube of 5]

We know that $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

Let, a = 3y, b= 5z

Hence,

$(3 y)^{3}+(5 z)^{3}=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$

$(3 y)^{3}+(5 z)^{3}=(3 y+5 z)\left(9 y^{2}+25 z^{2}-15 y z\right) \rightarrow(2)$

From eq. (1) and (2) we get,

$27 y^{3}+125 z^{3} \equiv(3 y+5 z)\left(9 y^{2}+25 z^{2}-15 y z\right)$

Answer 2

Answer:

$Factors\:of \: 27y^{3}+125z^{3}\\=(3y+5z)(9y^{2}-15yz+25z^{2})$

Step-by-step explanation:

$Given \: 27y^{3}+125z^{3}\\=3^{3}y^{3}+5^{3}z^{3}\\=\big(3y\big)^{3}+\big(5z\big)^{3}$

$=(3y+5z)[(3y)^{2}-(3y)(5z)+(5z)^{2}]$

/* By Algebraic identity:

a³+b³ = (a+b)(a²-ab+b²) */

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

Therefore,

$Factors\:of \: 27y^{3}+125z^{3}\\=(3y+5z)(9y^{2}-15yz+25z^{2})$

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