Factorise each of the following: 27 p^{3} - f | vedclass

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Question

(A) The given expression is $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$.We can rewrite this expression as:$(3p)^{3} - (\frac{1}{6})

Vedclass · Accepted Answer

$(3p - 1/6)^3$

Vedclass · Answer

(A) The given expression is $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$.We can rewrite this expression as:$(3p)^{3} - (\frac{1}{6})^{3} - 3(3p)(\frac{1}{6})(3p - \frac{1}{6})$.This is in the form of the algebraic identity $x^{3} - y^{3} - 3xy(x - y) = (x - y)^{3}$,where $x = 3p$ and $y = \frac{1}{6}$.Therefore,the expression simplifies to $(3p - \frac{1}{6})^{3}$.This can be written as $(3p - \frac{1}{6})(3p - \frac{1}{6})(3p - \frac{1}{6})$.

Factorise each of the following: $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

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