+0

0
585
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
#1
+8

Which as you said gives:

Your two roots are given by

and

Therefore

Finally:

Guest Jun 8, 2015
#2
+20711
+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
+94202
+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
+94202
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
+94202
+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
+20711
+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