+0

0
433
7
+262

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

#1
+8

Which as you said gives:

Your two roots are given by

and

Therefore

Finally:

Guest Jun 8, 2015
Sort:

#1
+8

Which as you said gives:

Your two roots are given by

and

Therefore

Finally:

Guest Jun 8, 2015
#2
+19207
+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
+92254
+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
+92254
0

Hi anon,

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
+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
+92254
+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
+19207
+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

### 23 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details