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algebra intermediate

Problem

Compute

\frac{x ^{6} - 16 x ^{3} + 64}{x ^{3} - 8}

when

x = 6

.

Solution

Note that

(x^{3} - 8)^{2} = x^{6} - 16 x^{3} + 64

. So

\frac{x ^{6} - 16 x ^{3} + 64}{x ^{3} - 8} = \frac{( x ^{3} - 8 ) ^{2}}{x ^{3} - 8} = x^{3} - 8

. So, the answer is

6^{3} - 8 = 216 - 8 = 208

.

Final answer

208