Solve for y square root of y^6-2y^3+1=y^3-1

$\sqrt{y^{6} - 2 y^{3} + 1} = y^{3} - 1$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{y^{6} - 2 y^{3} + 1}}^{2} = {(y^{3} - 1)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y^{6} - 2 y^{3} + 1}$ as ${(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2}}$ .

${({(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2}})}^{2} = {(y^{3} - 1)}^{2}$

Simplify the left side.

Simplify ${({(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2}})}^{2}$ .

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

${(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2} \cdot 2} = {(y^{3} - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y^{6} - 2 y^{3} + 1)}^{\frac{1}{2} \cdot 2} = {(y^{3} - 1)}^{2}$

Rewrite the expression.

${(y^{6} - 2 y^{3} + 1)}^{1} = {(y^{3} - 1)}^{2}$

Simplify.

$y^{6} - 2 y^{3} + 1 = {(y^{3} - 1)}^{2}$

Simplify the right side.

Simplify ${(y^{3} - 1)}^{2}$ .

Rewrite ${(y^{3} - 1)}^{2}$ as $(y^{3} - 1) (y^{3} - 1)$ .

$y^{6} - 2 y^{3} + 1 = (y^{3} - 1) (y^{3} - 1)$

Expand $(y^{3} - 1) (y^{3} - 1)$ using the FOIL Method.

Apply the distributive property.

$y^{6} - 2 y^{3} + 1 = y^{3} (y^{3} - 1) - 1 (y^{3} - 1)$

Apply the distributive property.

$y^{6} - 2 y^{3} + 1 = y^{3} y^{3} + y^{3} \cdot - 1 - 1 (y^{3} - 1)$

Apply the distributive property.

$y^{6} - 2 y^{3} + 1 = y^{3} y^{3} + y^{3} \cdot - 1 - 1 y^{3} - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $y^{3}$ by $y^{3}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y^{6} - 2 y^{3} + 1 = y^{3 + 3} + y^{3} \cdot - 1 - 1 y^{3} - 1 \cdot - 1$

Add $3$ and $3$ .

$y^{6} - 2 y^{3} + 1 = y^{6} + y^{3} \cdot - 1 - 1 y^{3} - 1 \cdot - 1$

Move $- 1$ to the left of $y^{3}$ .

$y^{6} - 2 y^{3} + 1 = y^{6} - 1 \cdot y^{3} - 1 y^{3} - 1 \cdot - 1$

Rewrite $- 1 y^{3}$ as $- y^{3}$ .

$y^{6} - 2 y^{3} + 1 = y^{6} - y^{3} - 1 y^{3} - 1 \cdot - 1$

Rewrite $- 1 y^{3}$ as $- y^{3}$ .

$y^{6} - 2 y^{3} + 1 = y^{6} - y^{3} - y^{3} - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$y^{6} - 2 y^{3} + 1 = y^{6} - y^{3} - y^{3} + 1$

Subtract $y^{3}$ from $- y^{3}$ .

$y^{6} - 2 y^{3} + 1 = y^{6} - 2 y^{3} + 1$

Solve for $y$ .

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

Subtract $y^{6}$ from both sides of the equation.

$y^{6} - 2 y^{3} + 1 - y^{6} = - 2 y^{3} + 1$

Add $2 y^{3}$ to both sides of the equation.

$y^{6} - 2 y^{3} + 1 - y^{6} + 2 y^{3} = 1$

Combine the opposite terms in $y^{6} - 2 y^{3} + 1 - y^{6} + 2 y^{3}$ .

Subtract $y^{6}$ from $y^{6}$ .

$- 2 y^{3} + 1 + 0 + 2 y^{3} = 1$

Add $- 2 y^{3} + 1$ and $0$ .

$- 2 y^{3} + 1 + 2 y^{3} = 1$

Add $- 2 y^{3}$ and $2 y^{3}$ .

$0 + 1 = 1$

Add $0$ and $1$ .

$1 = 1$

Since $1 = 1$ , the equation will always be true for any value of $y$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for y square root of y^6-2y^3+1=y^3-1

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