need a help please.
that's what i did:
(x^2+2x)^3+8x^3=0
x^6+6x^5+12x^4+8x^3=0
x^3(x^3+6x^2+12x+8)=0
x^3=0 -> x=0
now what to do with the brackets ?
Which as you said gives:
Your two roots are given by
and
Therefore
Finally:
A:
$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left( x^3+3x^2\cdot 2 + 3x\cdot 2^2 + 2^3 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 +8\right) &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 16 \right) &=& 0\\
\end{array}
$}}$$
I. First solution
$$x^3=0 \qquad \Rightarrow \qquad x=0$$
II. Second solution
$$\underbrace{x^3+6x^2 + 12x + 16}_{=(x+4)(x^2+2x+4)} = 0 \qquad \Rightarrow \qquad x = -4$$
III. No more solutions
$$x^2+2x+4 = 0$$
The parabola has no solutions, because the vertex of this parabola (-1 | 3) lies above the x-axis
B:
$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left[~~\left( x+2 \right)^3 + 8~~\right] &=& 0\\
\end{array}
$}}$$
I. First solution:
$$x^3=0 \qquad \Rightarrow \qquad x=0$$
II. Second solution:
$$\small{\text{$
\begin{array}{rcl}
\left( x+2 \right)^3 + 8 &=& 0\\
\left( x+2 \right)^3 &=& -8 \qquad | \qquad \sqrt[3]{}\\
x+2 &=& \sqrt[3]{-8}\\
x+2 &=& -2\\
x &=& -2-2\\
x &=& -4\\
\end{array}
$}}$$
Thanks Heureka,
I would have done this as the sum of 2 cubes
$$\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\\\\\
(x(x+2))^3+8x^3=0\\\\
x^3((x+2)^3+8)=0\\\\
x=0\qquad or\\\\
(x+2)^3+2^3=0\\\\
((x+2)+2)((x+2)^2-2(x+2)+2^2)=0\\\\
(x+4)(x^2+4x+4-2x-4+4)=0\\\\$$
$$\\(x+4)(x^2+2x+4)=0\\\\
x=-4\qquad or\\\\
x^2+2x+4=0\\\\
$This has no real roots. \\\\So$\\\\
x=0 \;\;or\;\; x=-4\\\\$$
ACTUALLY, NOW I LOOK AT IT, HEUREKA'S ANSWER IS BETTER :)
Hi anon,
Thankyou for trying to answer.
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Why don't you join up. It is super easy, you just need to choose a username and a passsword :)
heureka i still dont understand the brackets part
why :x^3+6x^2+12x+16 -> (x+4)(x^2+2x+4)
Heureka's Part B method is much easier to follow.
Heureka would have got that factorisation by algebraic division.
heureka i still dont understand the brackets part
why :x^3+6x^2+12x+16 -> (x+4)(x^2+2x+4)
Hallo sabi92,
I. $$\small{\text{$
x^3+6x^2+12x+16:
$}}$$
$$\small{\text{$
x = -4 \qquad \Rightarrow \qquad (-4)^3+6\cdot (-4)^2 + 12\cdot (-4) + 16
= -64 + 96 -48 + 16 = 0
$}}$$
II. algebraic division:
$$\small{\text{$
x^3+6x^2+12x+16~~:~~(x+4) =x^2+2x+4
$}}$$