Factorise the following:8 p^{3}+frac{12}{5} p | vedclass

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(A) The given expression is $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$.We can rewrite this expression in the form of the algebraic iden

Vedclass · Accepted Answer

$(2p + \frac{1}{5})^3$

Vedclass · Answer

(A) The given expression is $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$.We can rewrite this expression in the form of the algebraic identity $(a+b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$.Here,$a = 2p$ and $b = \frac{1}{5}$.Substituting these values into the identity:$(2p)^{3} + 3(2p)^{2}(\frac{1}{5}) + 3(2p)(\frac{1}{5})^{2} + (\frac{1}{5})^{3}$$= 8p^{3} + 3(4p^{2})(\frac{1}{5}) + 3(2p)(\frac{1}{25}) + \frac{1}{125}$$= 8p^{3} + \frac{12}{5}p^{2} + \frac{6}{25}p + \frac{1}{125}$.Thus,the expression is equivalent to $(2p + \frac{1}{5})^{3}$,which can be written as $(2p + \frac{1}{5})(2p + \frac{1}{5})(2p + \frac{1}{5})$.

Factorise the following:
$8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$

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Factorise the following:$8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$