Solve for a (a^2+15a)^(1/3)=3(a-1)^(1/3)

${(a^{2} + 15 a)}^{\frac{1}{3}} = 3 {(a - 1)}^{\frac{1}{3}}$

Step 1

Eliminate the fractional exponents by multiplying both exponents by the LCD.

${({(a^{2} + 15 a)}^{\frac{1}{3}})}^{3} = {(3 {(a - 1)}^{\frac{1}{3}})}^{3}$

Step 2

Simplify ${({(a^{2} + 15 a)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(a^{2} + 15 a)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(a^{2} + 15 a)}^{\frac{1}{3} \cdot 3} = {(3 {(a - 1)}^{\frac{1}{3}})}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(a^{2} + 15 a)}^{\frac{1}{3} \cdot 3} = {(3 {(a - 1)}^{\frac{1}{3}})}^{3}$

Rewrite the expression.

${(a^{2} + 15 a)}^{1} = {(3 {(a - 1)}^{\frac{1}{3}})}^{3}$

Simplify.

$a^{2} + 15 a = {(3 {(a - 1)}^{\frac{1}{3}})}^{3}$

Step 3

Simplify ${(3 {(a - 1)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $3 {(a - 1)}^{\frac{1}{3}}$ .

$a^{2} + 15 a = 3^{3} {({(a - 1)}^{\frac{1}{3}})}^{3}$

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

$a^{2} + 15 a = 27 {({(a - 1)}^{\frac{1}{3}})}^{3}$

Multiply the exponents in ${({(a - 1)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{2} + 15 a = 27 {(a - 1)}^{\frac{1}{3} \cdot 3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$a^{2} + 15 a = 27 {(a - 1)}^{\frac{1}{3} \cdot 3}$

Rewrite the expression.

$a^{2} + 15 a = 27 {(a - 1)}^{1}$

Simplify.

$a^{2} + 15 a = 27 (a - 1)$

Apply the distributive property.

$a^{2} + 15 a = 27 a + 27 \cdot - 1$

Multiply $27$ by $- 1$ .

$a^{2} + 15 a = 27 a - 27$

Step 4

Move all terms containing $a$ to the left side of the equation.

Subtract $27 a$ from both sides of the equation.

$a^{2} + 15 a - 27 a = - 27$

Subtract $27 a$ from $15 a$ .

$a^{2} - 12 a = - 27$

Step 5

Add $27$ to both sides of the equation.

$a^{2} - 12 a + 27 = 0$

Step 6

Factor $a^{2} - 12 a + 27$ using the AC method.

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 $27$ and whose sum is $- 12$ .

$- 9, - 3$

Write the factored form using these integers.

$(a - 9) (a - 3) = 0$

Step 7

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

$a - 9 = 0$

$a - 3 = 0$

Step 8

Set $a - 9$ equal to $0$ and solve for $a$ .

Set $a - 9$ equal to $0$ .

$a - 9 = 0$

Add $9$ to both sides of the equation.

$a = 9$

Step 9

Set $a - 3$ equal to $0$ and solve for $a$ .

Set $a - 3$ equal to $0$ .

$a - 3 = 0$

Add $3$ to both sides of the equation.

$a = 3$

Step 10

The final solution is all the values that make $(a - 9) (a - 3) = 0$ true.

$a = 9, 3$

Solve for a (a^2+15a)^(1/3)=3(a-1)^(1/3)

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