1 Answer

Grant Langseth · Answer 1 · 2021-09-23T21:00:14+0000

Answer:

x = 18

Explanation:

Given
$3 \sqrt[4]{(x-2)^3}-4 = 20$

now adding '4' on both sides, we get

$3 \sqrt[4]{(x-2)^3}-4+4= 20+4$

$3 \sqrt[4]{(x-2)^3} = 24$

i)
$3 \sqrt[4]{(x-2)^3}$ = 24

cancelling '3' on both sides, we get

$\sqrt[4]{(x-2)^3} = 8$

ii)
$\sqrt[4]{(x-2)^3}$ = 8

iii)
$(x-2)^{(3)/(4) } = 8$

$(x-2)^{(3)/(4) } = 2^(3)$

iv)
$(x-2)^(3)/(4) )^{(1)/(3) } = (2^(3))^(1)/(3)$

now simplify on both sides,we get

(x-2)^(1)/(4) = 2

$((x-2)^{(1)/(4) })^4 = 2^4$

simplify we get solution is

v) (x-2) = 16

vi) final answer is x = 16 +2 =18

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