Question

`\huge\text{Please help!}`

Simplify:

`\bold{(x^(2) +8xy+16y^(2))^{(1)/(3) } (x+4y)}^{(1)/(3) }`
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`\bold{M^ag^ic^al\:^}`

Answer 1

`\bold{x + 4y}`

Explanation :-

`{(x^(2) +8xy+16y^(2))^{(1)/(3) } (x+4y)}^{(1)/(3) }`

Rewriting it into the expression of
`a^(2) + 2ab + {b}^(2)`,

`x^(2) + 2x * 4y + (4y)^(2))^{(1)/(3) }(x + 4y)^{ (1)/(3) }`

`= > ((x + 4y)^(2))^{ (1)/(3) }(x + 4y)^{ (1)/(3) }`

Now multiplying the fractions as,

`= > ((x + 4y))^{ (2)/(3) }(x + 4y)^{ (1)/(3) }`

Now as the bases of the fractions are same, we can multiply directly,

`= > (x + 4y)^{ (2)/(3) + (1)/(3) }`

`= > (x + 4y)^{ (3)/(3) }`

As when same numerator and denominator of a fraction is 1, so,

`= > (x + 4y)^1`

As 1 can be neglected (since it has no specific need in the equation), so,

`= > x + 4y`

Hence, the answer is
`\bold{x + 4y}`

Answer 2

x + 4y

Explanation:

Hey there! First, we have to recall back laws of exponent.

`\displaystyle \large{a^{(m)/(n)} = \sqrt[n]{a^m} }`

Now simplify the expressions in surds form.

`\displaystyle \large{\sqrt[3]{x^2+8xy+16y^2} \cdot \sqrt[3]{x+4y} }`

From x²+8xy+16y², we can convert the expression to perfect square.

Perfect Square

`\displaystyle \large{x^2+2xb+b^2 = (x+b)^2}`

Therefore, from the expression.

`\displaystyle \large{x^2+8xy+16y^2 = (x+4y)^2}`

Thus:

`\displaystyle \large{\sqrt[3]{(x+4y)^2} \cdot \sqrt[3]{x+4y}`

Because both have same surds, multiply them in one.

Surd Property I

`\displaystyle \large{\sqrt[n]{x} \cdot \sqrt[n]{y} =\sqrt[n]{xy} }`

Therefore:

`\displaystyle \large{\sqrt[3]{(x+4y)^2(x+4y)}`

Since both are like-terms and multiplying each other, we can apply one of exponent laws.

Exponent Laws II

`\displaystyle \large{a^m \cdot a^n = a^(m+n)}`

Therefore, we have
`\displaystyle \large{\sqrt[3]{(x+4y)^(2+1) } \to \sqrt[3]{(x+4y)^3} }`

Simplify the expression, we have cube expression inside the cube root. Therefore, we simplify as x+4y as we cancel cube and cube root.

Let me know if you have any questions through comments!