\tt If\;\; \sqrt[3]{x}+a=b,\:then \; x=

Question

`\tt If\;\; \sqrt[3]{x}+a=b,\:then \; x=`

Answer 1

x = b³ - 3b²a + 3a²b - a³

Explanation:

Given:

`\sqrt[3]{x} + a = b`

To determine the value of "x", we need to isolate it. Start out by subtracting "a" both sides. Then, we can cube both sides to remove the cube root. Finally, we can simplify both sides to obtain the value of x.

Subtract "a" both sides to isolate ∛x:

`\implies \sqrt[3]{x} + a - a = b - a`

`\implies \sqrt[3]{x} = b - a`

Cube both sides to remove the cube root on the L.H.S:

`\implies (\sqrt[3]{x})^(3) = (b - a)^(3)`

`\implies x= (b - a)^(3)`

Use the formula (a - b)³ = a³ - 3a²b + 3ab² - b³

`\implies x = b^(3) - 3(b)^(2)(a) + 3(b)(a)^(2) - a^(3)`

`\implies \boxed{x = b^(3) - 3b^(2)a + 3a^(2)b - a^(3)}`

Thus, the value of x is b³ - 3b²a + 3a²b - a³.

Answer 2

`x=(b-a)^(3)`

Explanation:

• Subtract ‘a’ from both sides
• This gives
  `\sqrt[3]{x}=b-a`
• Cube both sides
• This gives
  `x=(b-a)^(3)`