Question

Factorise 27p3-1/216-9/2p2+1/4p

Answer 1

=(3p)³ - (1/6)³ - 3×(3p)²×1/6 +3×3p×(1/6)²
=(3p - 1/6)³
=(3p-1/6)(3p-1/6)(3p-1/6)

Answer 2

Answer:

factorization of $27p^3-\frac{1}{216} -\frac{9}{2}p^2+\frac{1}{4}p$ is $(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$

Step-by-step explanation:

We have to factorize

$27p^3-\frac{1}{216} -\frac{9}{2}p^2+\frac{1}{4}p$

We know ,

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

given expression can be written as,

$(3p)^3-(\frac{1}{6})^3 -\frac{9}{2}p^2+\frac{1}{4}p$

Comparing we get,

a= 3p and $b=\frac{1}{6}$

Then , expression becomes,

$(3p)^3-(\frac{1}{6})^3 -\frac{9}{2}p^2+\frac{1}{4}p=(3p-\frac{1}{6})^3$

In factored form it can be written as ,

$(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$

Thus,factorization of $27p^3-\frac{1}{216} -\frac{9}{2}p^2+\frac{1}{4}p$ is $(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$