1. Solving Cubic Equations

Hi,

Could you please explain how to solve these two sums, step by step please.

1. $\displaystyle 2x^3=16x$
2. $\displaystyle 2(x-1)^3=32$

This is as far as I got on both sums:

1.
$\displaystyle 2x^3=16x$
$\displaystyle x^3=8x$
$\displaystyle x^2=8$
$\displaystyle x^2-8=0$

2.
$\displaystyle 2(x-1)^3=32$
$\displaystyle (x-1)^3=16$
$\displaystyle (x-1)^3-16=0$

Any help would be greatly appreciated,

Thanks,

Dru

2. Hi.

Originally Posted by 22upon7
Hi,

Could you please explain how to solve these two sums, step by step please.

1. $\displaystyle 2x^3=16x$
2. $\displaystyle 2(x-1)^3=32$

This is as far as I got on both sums:

1.
$\displaystyle 2x^3=16x$
$\displaystyle x^3=8x$
$\displaystyle x^2=8$
$\displaystyle x^2-8=0$
Good thinking, but you made a mistake

$\displaystyle x^3=8x$

OKAY

$\displaystyle x^2=8$

Not okay, you divided by x; you can do that, but this is only for $\displaystyle x \not= 0$; and x = 0 is a solution!

So

$\displaystyle x^3=8x$

$\displaystyle x^3-8x = 0$

$\displaystyle x(x^2-8) = 0$

is better,

now you know x = 0 is a solution and x²-8 = 0 leads to the other solutions; you still have to solve x²-8= 0, tho.

Originally Posted by 22upon7
2.
$\displaystyle 2(x-1)^3=32$
$\displaystyle (x-1)^3=16$
$\displaystyle (x-1)^3-16=0$
The last step seems to be unnecessary

$\displaystyle (x-1)^3=16$ // ^(1/3)

$\displaystyle x-1 = 16^{1/3}$

$\displaystyle x = 16^{1/3}+1$

Originally Posted by 22upon7
Any help would be greatly appreciated,

Thanks,
You're welcome.

Yours, Rapha