Question

Factorise 27x^3-1/216-9/2x^2+1/4x

Answer 1

Answer:

27x∧3-1/216-9/2x∧2+1/4x

    27x∧3-9/2x2+1/4x-1/216

     (3x)∧3-3(3x)∧2*1/6+3*3x*(1/6)∧2-(1/6)∧3

     (3x-1/6)∧3                     ∵(a-b)∧3

Step-by-step explanation:

Answer 2

Given,

• $27x^3 -1/216 -9/2x^2 +1/4x$ is given.

To find,

• We have to find the factors of the given cubic equation.

Solution,

The factors of $27x^3 -1/216 -9/2x^2 +1/4x$ are $(3x-1/6)(3x-1/6)(3x-1/6)$.

We can simply factorize the given cubic equation $27x^3 -1/216 -9/2x^2 +1/4x$ by using the algebraic identity,

$(a-b)^3 = a^3 -b^3 - 3a^2b +3ab^2$     (1)

where a = 3x, b = 1/6

Substituting the values of a and b in (1), we get

$(3x-1/6)^3 = 27x^3 -1/216 -9/2x^2 +1/4x$

∴ The factors of $27x^3 -1/216 -9/2x^2 +1/4x$ are $(3x-1/6)(3x-1/6)(3x-1/6)$.

Hence, the factors of $27x^3 -1/216 -9/2x^2 +1/4x$are $(3x-1/6)(3x-1/6)(3x-1/6)$.