Question 55

$$\frac{x^2(x - 4)^2}{(x + 4)^2 - 4x} \div \frac{(x^2 - 4x)^3}{(x + 4)^2} \times \frac{64 - x^3}{16 - x^2}$$ is equal to

Solution

$$\frac{x^2(x - 4)^2}{(x + 4)^2 - 4x} \div \frac{(x^2 - 4x)^3}{(x + 4)^2} \times \frac{64 - x^3}{16 - x^2}$$

= $$\frac{x^2(x - 4)^2}{(x + 4)^2 - 4x} \times \frac{(x + 4)^2}{(x^2 - 4x)^3} \times \frac{64 - x^3}{16 - x^2}$$

= $$\frac{x^2(x - 4)^2}{(x + 4)^2 - 4x} \times \frac{(x + 4)^2}{x^3(x - 4)^3} \times \frac{4^3 - x^3}{4^2 - x^2}$$

= $$\frac{1}{(x + 4)^2 - 4x} \times \frac{(x + 4)^2}{x(x - 4)} \times \frac{(4 - x)(4^2 + 4x + x^2)}{(4 - x)(4 + x)}$$

= $$\frac{1}{4^2 + 8x + x^2 - 4x} \times \frac{(x + 4)}x(x - 4)} \times (4^2 + 4x + x^2)$$

= $$\frac{1}{4^2 + 4x + x^2} \times \frac{(x + 4)}{x(x - 4)} \times (4^2 + 4x + x^2)$$

= $$\frac{(x + 4)}{x(x - 4)} $$


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