Question

16) Factorize 1) 216 p3-1/343-108/7 p2+18/49 p​

Answer 1

Answer:

$216p^{3}-\frac{1}{343}-\frac{108}{7}p^{2}+\frac{18}{49}p\\=(6p-\frac{1}{7})(6p-\frac{1}{7})(6p-\frac{1}{7})$

Step-by-step explanation:

$Given \: 216p^{3}-\frac{1}{343}-\frac{108}{7}p^{2}+\frac{18}{49}p$

$=216p^{3}-\frac{108}{7}p+\frac{18}{49}p-\frac{1}{343}$

$=(6p)^{3}-3\times (6p)^{2}\times \frac{1}{7}+3\times (6p)\times \left(\frac{1}{7}\right)^{2}-\left(\frac{1}{7}\right)^{3}$

$=(6p-\frac{1}{7})^{3}$

/* By algebraic identity:

a³-3a²b+3ab²-b³ = (a-b)³ */

Therefore,

$216p^{3}-\frac{1}{343}-\frac{108}{7}p^{2}+\frac{18}{49}p\\=(6p-\frac{1}{7})(6p-\frac{1}{7})(6p-\frac{1}{7})$

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