Question

plz solve.............

Answer 1

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

$(2p)^3 + 3\cdot(2p)^2\cdot\frac{1}{5} + 3\cdot2p\cdot(\frac{1}{5})^2 + (\frac{1}{5})^3$

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

Answer 2

Hope u like my process
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Formula to be used
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$= > \boxed{ \it \orange{ {x}^{3} + 3 {x}^{2}y + 3x {y}^{2} + {y}^{3}} = \orange{(x + y) {}^{3} } }$
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Now,..
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$= > \bf \blue{8 {p}^{3} + \frac{12}{5} {p}^{2} + \frac{6}{25} p + \frac{1}{125} } \\ \\ = \blue{ {(2p)}^{3} + 3 {(2)}^{2p} ( \frac{1}{5} ) + 3(2) {( \frac{1}{5}p )}^{2} + {( \frac{1}{5} )}^{3} } \\ \\ = \boxed{ \bf \underline{\blue{ {(2p + \frac{1}{5} )}^{3} }}}$
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Alternative process
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$= > \green{ \bf \: 8 {p}^{3} + \frac{12}{5} {p}^{2} + \frac{6}{25} p + \frac{1}{125} } \\ \\ = \green{8 {p}^{3} + \frac{4}{5} {p}^{2} + \frac{8}{5} {p}^{2} + \frac{4}{25}p + \frac{2}{25}p + \frac{1}{125} } \\ \\ = \green{4 {p}^{2} (2p + \frac{1}{5} ) + \frac{4}{5}p (2p + \frac{1}{5}) + \frac{1}{25}(2p + \frac{1}{5} ) } \\ \\ = \green{(2p + \frac{1}{5})(4 {p}^{2} + \frac{4}{5} p + \frac{1}{25} ) } \\ \\ = \green{ (2p + \frac{1}{5} )(4 {p}^{2} + \frac{2}{5} p + \frac{2}{5}p + \frac{1}{25} ) } \\ \\ = \green{(2p + \frac{1}{5})(2p(2p + \frac{1}{5} ) + \frac{1}{5} (2p + \frac{1}{5} )) } \\ \\ = \green{(2p + \frac{1}{5} )(2p + \frac{1}{5} )(2p + \frac{1}{5} )} \\ \\ = \boxed{ \underline{ \green{ {(2p + \frac{1}{5} )}^{3} }}}$
Thus the required factorization is.

$= > \boxed{ \orange{ {(2p + \frac{1}{5}) }^{3} }}$
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Hope this is ur required answer

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