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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Factorise: $49 a^{2}+70 a b+25 b^{2}$

Verify whether the following is a zero of the polynomial indicated against it:
$p(x) = 3x + 1, \, x = -\frac{1}{3}$

Factorise $y^2 - 5y + 6$ by using the Factor Theorem.

Find $p(0)$,$p(1)$,and $p(2)$ for the following polynomial: $p(t) = 2 + t + 2t^2 - t^3$.

Verify whether the following is a zero of the polynomial indicated against it:
$p(x) = 2x + 1, \, x = \frac{1}{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module