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Quantitative Aptitude Question on Algebra

What is the value of 'x' in the expression given below ?
x4−16x2−6x+9×x3−27(x2+4)(x+2)÷x2+3x+9x−3=32\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}

A

2.5

B

3.5

C

1.5

D

4.5

Answer

3.5

Explanation

Solution

We know, (a - b)2 = a2 - 2ab + b2
a3 - b3 = (a - b)(a2 + ab - b2)
(a4 - b4) = (a2 + b2)(a + b)(a - b)
x4−16x2−6x+9×x3−27(x2+4)(x+2)÷x2+3x+9x−3=32\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}
Now, (x2+4)(x−2)(x+2)(x−3)(x−3)×(x−3)(x2+3x+9)(x2+4)(x+2)×x−3x2+3x+9=32\frac{(x^2+4)(x-2)(x+2)}{(x-3)(x-3)}\times\frac{(x-3)(x^2+3x+9)}{(x^2+4)(x+2)}\times\frac{x-3}{x^2+3x+9}=\frac{3}{2}
So, the correct option is (B) : 3.5