Question

Express the rational expression x²-4/x³+8 in its lowest form.

Answer 1

Step-by-step explanation:

Remember The Identity :

${x}^{2} - {y}^{2} = (x + y)(x - y) \\ \\ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

Here's The Answer My Friend....

Answer 2

The lowest form of expression is $=\frac{x-2}{x^{2} -2x +4}$.

Given: $\frac{x^{2-4} }{x^{3}+8 }$

To find: the expression in its lowest form

Solution:

Taking the formulas,

$a^{2}$ - $b^{2}$ = (a + b) (a - b)

$a^{3}$ + $b^{3}$ = $(a+b)^{3}$ - 3ab (a + b)

$\frac{x^{2-4} }{x^{3}+8 }$

$= \frac{x^{2} - 2^{2} }{x^{3} + 2^{3} }$

= $\frac{(x + 2) (x-2)}{(x + 2)^{3} - 3\times x\times2 (x+2)}$

$=\frac{(x+2)(x-2)}{(x+2)[(x+2)^{2}-6x] }$

(x + 2) gets canceled and we get,

$= \frac{x-2}{(x+2)^{2}-6x }$

$=\frac{x-2}{x^{2} +4x +4 -6x}$

$=\frac{x-2}{x^{2} -2x +4}$

Answer: The lowest form of the expression is $=\frac{x-2}{x^{2} -2x +4}$.

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