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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 ? 

sabi92  Jun 8, 2015

Best Answer 

 #1
avatar
+8

 

(x(x+2))^3 +8x^3=0

x^3(x+2)^3 +8x^3 =0

 

Which as you said gives:

x^3((x+2)^3 +8)=0

Your two roots are given by

x^3 =0 \to x=0

and

(x+2)^3 +8 =0

Therefore

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

Finally:

x=-2-2 = -4

Guest Jun 8, 2015
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7+0 Answers

 #1
avatar
+8
Best Answer

 

(x(x+2))^3 +8x^3=0

x^3(x+2)^3 +8x^3 =0

 

Which as you said gives:

x^3((x+2)^3 +8)=0

Your two roots are given by

x^3 =0 \to x=0

and

(x+2)^3 +8 =0

Therefore

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

Finally:

x=-2-2 = -4

Guest Jun 8, 2015
 #2
avatar+18712 
+5

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:

$$\small{\text{$
\begin{array}{rcl}
\left( x+2 \right)^3 + 8 &=& 0\\
\left( x+2 \right)^3 &=& -8 \qquad | \qquad \sqrt[3]{}\\
x+2 &=& \sqrt[3]{-8}\\
x+2 &=& -2\\
x &=& -2-2\\
x &=& -4\\
\end{array}
$}}$$

heureka  Jun 8, 2015
 #3
avatar+90988 
+5

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 :)

Melody  Jun 8, 2015
 #4
avatar+90988 
0

Hi anon,

Thankyou for trying to answer.

You can no longer post pictures unless you are a member.

Why don't you join up.  It is super easy, you just need to choose a username and a passsword :)

Melody  Jun 8, 2015
 #5
avatar+262 
+5

heureka i still dont understand the brackets part

why :x^3+6x^2+12x+16  -> (x+4)(x^2+2x+4)

sabi92  Jun 8, 2015
 #6
avatar+90988 
+5

Heureka's Part B method is much easier to follow.

 

 

Heureka would have got that factorisation by algebraic division.    

Melody  Jun 8, 2015
 #7
avatar+18712 
+5

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:

$$\small{\text{$
x^3+6x^2+12x+16~~:~~(x+4) =x^2+2x+4
$}}$$

heureka  Jun 9, 2015

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