Question

10. If x‐² = 64, then x⅓ + x⁰ =
(a) 2
(b) 3
(c) 3/2
(d) 2/3​

Answer 1

$\huge \fbox \red{Answer : Option (C) }$

Solution :

Given that :

${x}^{ - 2} = 64 \\ \\ = > \frac{1}{ {x}^{2} } = 64 \\ \\ = > {x}^{2} = \frac{1}{64} \\ \\ = > x = \sqrt{ \frac{1}{64} } = \frac{1}{8}$

Then,

${x}^{ \frac{1}{3} } + {x}^{0} \\ \\ = > { (\frac{1}{8}) }^{ \frac{1}{3} } + 1 \\ \\ = > \frac{1}{ {8}^{ \frac{1}{3} } } + 1 \\ \\ = > \frac{1}{ { ({2}^{3} )}^{ \frac{1}{3} } } + 1 \\ \\ = > \frac{1}{ {2}^{3 \times \frac{1}{3} } } + 1 = \frac{1}{2} + 1 = \frac{3}{2}$

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Hope it helps ☺

Fóllòw Më ❤