Solve \(1 + \sqrt[3]{ x - 3} = 4\)
\(1 + \sqrt [3]{x-3} = 4\)
= \(\sqrt[3]{x - 3} = 4 - 1\)
\(\sqrt[3]{x - 3} = 3 \)
take the cube of both sides
= x - 3 = 27 x = 27 + 3 ∴ x = 30
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