Verify the Identity x^3-2=(x- cube root of 2)(x^2+ cube root of 2x+ cube root of 4)

$x^{3} - 2 = (x - \sqrt[3]{2}) (x^{2} + \sqrt[3]{2} x + \sqrt[3]{4})$

Simplify the right side.

Tap for more steps...

Simplify $(x - \sqrt[3]{2}) (x^{2} + \sqrt[3]{2} x + \sqrt[3]{4})$ .

Tap for more steps...

Expand $(x - \sqrt[3]{2}) (x^{2} + \sqrt[3]{2} x + \sqrt[3]{4})$ by multiplying each term in the first expression by each term in the second expression.

$x^{3} - 2 = x \cdot x^{2} + x (\sqrt[3]{2} x) + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Simplify terms.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $x$ by $x^{2}$ by adding the exponents.

Tap for more steps...

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Raise $x$ to the power of $1$ .

$x^{3} - 2 = x^{1} x^{2} + x (\sqrt[3]{2} x) + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

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

$x^{3} - 2 = x^{1 + 2} + x (\sqrt[3]{2} x) + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Add $1$ and $2$ .

$x^{3} - 2 = x^{3} + x (\sqrt[3]{2} x) + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Rewrite using the commutative property of multiplication.

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x \cdot x + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Move $x$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} (x \cdot x) + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Multiply $x$ by $x$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2} (\sqrt[3]{2} x) - \sqrt[3]{2} \sqrt[3]{4}$

Multiply $- \sqrt[3]{2} (\sqrt[3]{2} x)$ .

Tap for more steps...

Raise $\sqrt[3]{2}$ to the power of $1$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - ({\sqrt[3]{2}}^{1} \sqrt[3]{2}) x - \sqrt[3]{2} \sqrt[3]{4}$

Raise $\sqrt[3]{2}$ to the power of $1$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - ({\sqrt[3]{2}}^{1} {\sqrt[3]{2}}^{1}) x - \sqrt[3]{2} \sqrt[3]{4}$

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

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - {\sqrt[3]{2}}^{1 + 1} x - \sqrt[3]{2} \sqrt[3]{4}$

Add $1$ and $1$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - {\sqrt[3]{2}}^{2} x - \sqrt[3]{2} \sqrt[3]{4}$

Rewrite ${\sqrt[3]{2}}^{2}$ as ${(2^{2})}^{\frac{1}{3}}$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{2^{2}} x - \sqrt[3]{2} \sqrt[3]{4}$

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

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - \sqrt[3]{2} \sqrt[3]{4}$

Multiply $- \sqrt[3]{2} \sqrt[3]{4}$ .

Tap for more steps...

Combine using the product rule for radicals.

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - \sqrt[3]{4 \cdot 2}$

Multiply $4$ by $2$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - \sqrt[3]{8}$

Rewrite $8$ as $2^{3}$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - \sqrt[3]{2^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$x^{3} - 2 = x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - 2$

Combine the opposite terms in $x^{3} + \sqrt[3]{2} x^{2} + x \sqrt[3]{4} - \sqrt[3]{2} x^{2} - \sqrt[3]{4} x - 2$ .

Tap for more steps...

Subtract $\sqrt[3]{2} x^{2}$ from $\sqrt[3]{2} x^{2}$ .

$x^{3} - 2 = x^{3} + x \sqrt[3]{4} + 0 - \sqrt[3]{4} x - 2$

Add $x^{3} + x \sqrt[3]{4}$ and $0$ .

$x^{3} - 2 = x^{3} + x \sqrt[3]{4} - \sqrt[3]{4} x - 2$

Reorder the factors in the terms $x \sqrt[3]{4}$ and $- \sqrt[3]{4} x$ .

$x^{3} - 2 = x^{3} + x \sqrt[3]{4} - x \sqrt[3]{4} - 2$

Subtract $x \sqrt[3]{4}$ from $x \sqrt[3]{4}$ .

$x^{3} - 2 = x^{3} + 0 - 2$

Add $x^{3}$ and $0$ .

$x^{3} - 2 = x^{3} - 2$

Since the two sides have been shown to be equivalent, the equation is an identity.

$x^{3} - 2 = (x - \sqrt[3]{2}) (x^{2} + \sqrt[3]{2} x + \sqrt[3]{4})$ is an identity.

Do you know how to Verify the Identity x^3-2=(x- cube root of 2)(x^2+ cube root of 2x+ cube root of 4)? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.