Question

CUBE ROOT OF 7XCUBE MINUS 8 IS EQ
$\sqrt[3]{7 {x}^{3 } - 8 = - x }$
Find x
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Answer 1

I think there is a mistake in your question ,

the right question should be ,

$\sqrt[3]{7 {x}^{3} - 8 } = - x \\ \\ {( \sqrt[3]{7 {x}^{3} - 8} )}^{3} = { - x}^{3} \\ \\ 7 {x}^{3} - 8 = - {x}^{3} \\ \\ 7 {x}^{3} + {x}^{3} = 8 \\ \\ 8 {x}^{3} = 8 \\ \\ {x}^{3} = 1 \\ \\ x = 1$

Answer 2

✤✤✤QUESTION✤✤✤

$\sqrt[3]{7 {x}^{3} - 8} = - x$

find x.

✤✤✤ANSWER✤✤✤

We have,

$\sqrt[3]{7 {x}^{3} - 8 } = - x \\ \\ (cubing \: both \: sides) \\ \\ ({ \sqrt[3]{7 {x}^{3} - 8}) }^{3} = {( - x)}^{3} \\ \\ 7 {x}^{3} - 8 = { - x}^{3} \\ \\ 7 {x}^{3} + {x}^{3} = 8 \\ \\ 8 {x}^{3} = 8 \\ \\ {x}^{3} = 1 \\ \\ x = \sqrt[3]{1} \\ \\ x = 1$

HENCE THE VALUE OF X IS 1.