# Sum or difference of cubes.

• Jan 14th 2010, 06:07 PM
integral
Sum or difference of cubes.
$a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})$
And just switch the sign after $a^3-1$ for $a^3-b^3$

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

now take this example:

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

The end signs are clearly not $+,+$
Am I missing something?(Wondering)
• Jan 14th 2010, 06:42 PM
Prove It
Quote:

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

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

now take this example:

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

The end signs are clearly not $+,+$
Am I missing something?(Wondering)

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

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

$= (2x - 1)(4x^2 + 2x + 1)$.
• Jan 14th 2010, 06:51 PM
integral
$a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})$

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

$8x^3-1=(2x-1)(4x^2-2x-1)
$
• Jan 14th 2010, 06:52 PM
Prove It
What you have at the moment is a DIFFERENCE of cubes.

Use the Difference of Two cubes rule.

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

See my post above.

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

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

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