Question

If $x =\sqrt{3} - 1$ and $y =\sqrt{3}+1$ then $\frac{(x^4-y^4)}{(x+y)^2}$ is equal to ?

• A. $$4\sqrt{3}{9}$$
• B. $${9}{4\sqrt{3}}$$
• C. $$\frac{-4\sqrt{3}}{3}$$
• D. $$\frac{-8\sqrt{3}}{3}$$

Answer 1

Answer:

$$\frac{-8\sqrt{3}}{3}$$

Solution:

Given:

x = √3 - 1

y = √3 + 1

Formula used:

(x⁴ - y⁴) = (x - y) (x + y) (x²+ y²)

$(x+y)^2=x^2+y^2+2xy$

Calculation:

$\frac{(x^4-y^4)}{(x=y)^2}=\frac{(x-y)(x+y)(x^2+y^2)}{x^2+y^2+2xy}$

⇒x⁴ - y⁴ = [√3 - 1 - (√3 + 1)][√3 - 1 + √3 + 1] [(√3 + 1)² + (√3 - 1)²]  --------(1)

⇒ (x + y)² = (√3 - 1)² + (√3 + 1)² + 2(√3 - 1)(√3 + 1)   ---------(2)

On solving equation (1) and (2) separately we get:

x⁴ - y⁴ = -32√3

(x + y)² = 12

⇒ $\frac{x^4-y^4}{(x+y)^2}=\frac{-32\sqrt{3}}{12}$

⇒ $\frac{-8\sqrt{3}}{3}$

∴ The correct answer is $\frac{-8\sqrt{3}}{3}$.