What is (x^3-8)/(x^2-4x+4) ?

The solution to (x^3-8)/(x^2-4x+4) is (x^2+2x+4)/(x-2)

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simplify (x^3-8)/(x^2-4x+4)

simplify

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

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

Solution steps

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

Factor

x^{3} - 8 : (x - 2) (x^{2} + 2 x + 4)

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

Factor

x^{2} - 4 x + 4 : (x - 2)^{2}

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(x - 2)^{2} = (x - 2) (x - 2)

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

Cancel the common factor:

x - 2

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

\frac{5}{6} \cdot \frac{3}{8}

\frac{2 x + 10}{x + 7 - 2}

(5 x + 7)^{8}

- 43

- 2 x \cdot (- x)

What is (x^3-8)/(x^2-4x+4) ?
The solution to (x^3-8)/(x^2-4x+4) is (x^2+2x+4)/(x-2)