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# Pls help

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If $$\left(a-\frac 1a\right)^2=4$$ and $$a>1$$ what is the absolute value of $$a^3 - \frac1{a^3}?$$

Sep 16, 2021

#1
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(a - 1/a) = 2 (note that normally we couldn't assume that it isn't -2, but since a > 1, a - 1/a has to be positive.

(a - 1/a)^3 = 8

a^3 - 3a + 3/a - 1/a^3 = 8

a^3 - 1/a^3 - 3(a - 1/a) = 8

Can you figure it out from here?

=^._.^=

Sep 17, 2021