Thread: Sum or difference of cubes.

1. Sum or difference of cubes.

$\displaystyle a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})$
And just switch the sign after $\displaystyle a^3-1$ for$\displaystyle a^3-b^3$

My question:
I have read that the last two signs are constant (never changing) Unless the initial sign is different. I.e.
$\displaystyle a-b = +,+$
$\displaystyle a+b = -,+$

now take this example:

$\displaystyle 8x^3-1$
when worked out it equals:
$\displaystyle (\sqrt [3] {8x^3} + \sqrt [3] {-1})(8x^{3-1}+(-1)2x+(-1^{3-1})$
$\displaystyle (2x-1)(4x^2-2x-1)$

The end signs are clearly not $\displaystyle +,+$
Am I missing something?

2. Originally Posted by integral
$\displaystyle a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})$
And just switch the sign after $\displaystyle a^3-1$ for$\displaystyle a^3-b^3$

My question:
I have read that the last two signs are constant (never changing) Unless the initial sign is different. I.e.
$\displaystyle a-b = +,+$
$\displaystyle a+b = -,+$

now take this example:

$\displaystyle 8x^3-1$
when worked out it equals:
$\displaystyle (\sqrt [3] {8x^3} + \sqrt [3] {-1})(8x^{3-1}+(-1)2x+(-1^{3-1})$
$\displaystyle (2x-1)(4x^2-2x-1)$

The end signs are clearly not $\displaystyle +,+$
Am I missing something?
$\displaystyle a^3 \pm b^3 = (a - b)(a^2 \mp ab + b^2)$.

$\displaystyle 8x^3 - 1 = (2x)^3 - (1)^3$

$\displaystyle = (2x - 1)(4x^2 + 2x + 1)$.

3. $\displaystyle a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})$

$\displaystyle 8x^3-1=( \sqrt[3]{8x^{3}}+ \sqrt[3]{-1^{3}}) (8x^{3-1} -\sqrt[3]{8x^3b^3} +(-1^{3-1}))$
$\displaystyle 8x^3-1=(2x-1)(4x^2+(-1)2x+(-1)$
$\displaystyle 8x^3-1=(2x-1)(4x^2-2x-1)$

4. What you have at the moment is a DIFFERENCE of cubes.

Use the Difference of Two cubes rule.

$\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

See my post above.

Edit: Also, $\displaystyle -\sqrt[3]{-1(8x^3)} = -\sqrt[3]{-8x^3} = -(-2x) = 2x$

and $\displaystyle + (-1)^2 = +(1) = 1$.

So you do get +, + as your final two signs.