Solve by Factoring x^(2/3)-7x^(1/3)+10=0

$x^{\frac{2}{3}} - 7 x^{\frac{1}{3}} + 10 = 0$

Rewrite $x^{\frac{2}{3}}$ as ${(x^{\frac{1}{3}})}^{2}$ .

${(x^{\frac{1}{3}})}^{2} - 7 x^{\frac{1}{3}} + 10 = 0$

Let $u = x^{\frac{1}{3}}$ . Substitute $u$ for all occurrences of $x^{\frac{1}{3}}$ .

$u^{2} - 7 u + 10 = 0$

Factor $u^{2} - 7 u + 10$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$(u - 5) (u - 2) = 0$

Replace all occurrences of $u$ with $x^{\frac{1}{3}}$ .

$(x^{\frac{1}{3}} - 5) (x^{\frac{1}{3}} - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{\frac{1}{3}} - 5 = 0$

$x^{\frac{1}{3}} - 2 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x^{\frac{1}{3}} - 5 = 0$

Add $5$ to both sides of the equation.

$x^{\frac{1}{3}} = 5$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(x^{\frac{1}{3}})}^{3} = {(5)}^{3}$

Raise $5$ to the power of $3$ .

$x = 125$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x^{\frac{1}{3}} - 2 = 0$

Add $2$ to both sides of the equation.

$x^{\frac{1}{3}} = 2$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(x^{\frac{1}{3}})}^{3} = {(2)}^{3}$

Raise $2$ to the power of $3$ .

$x = 8$

The final solution is all the values that make $(x^{\frac{1}{3}} - 5) (x^{\frac{1}{3}} - 2) = 0$ true.

$x = 125, 8$

Solve by Factoring x^(2/3)-7x^(1/3)+10=0