Question:
What is the value of 'x' in the expression given below ?
\(\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}\)
What is the value of 'x' in the expression given below ?
\(\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}\)
\(\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}\)
Updated On: Nov 17, 2025
- 2.5
- 3.5
- 1.5
- 4.5
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The Correct Option is B
Solution and Explanation
We know, (a - b)2 = a2 - 2ab + b2
a3 - b3 = (a - b)(a2 + ab - b2)
(a4 - b4) = (a2 + b2)(a + b)(a - b)
\(\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}\)
Now, \(\frac{(x^2+4)(x-2)(x+2)}{(x-3)(x-3)}\times\frac{(x-3)(x^2+3x+9)}{(x^2+4)(x+2)}\times\frac{x-3}{x^2+3x+9}=\frac{3}{2}\)
So, the correct option is (B) : 3.5
a3 - b3 = (a - b)(a2 + ab - b2)
(a4 - b4) = (a2 + b2)(a + b)(a - b)
\(\frac{x^4-16}{x^2-6x+9}\times\frac{x^3-27}{(x^2+4)(x+2)}\div\frac{x^2+3x+9}{x-3}=\frac{3}{2}\)
Now, \(\frac{(x^2+4)(x-2)(x+2)}{(x-3)(x-3)}\times\frac{(x-3)(x^2+3x+9)}{(x^2+4)(x+2)}\times\frac{x-3}{x^2+3x+9}=\frac{3}{2}\)
So, the correct option is (B) : 3.5
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