Solve Equation: (7x+1)^(1/3)+(-x^2+x+8)^(1/3)+(x^2-8x-1)^(1/3)=2

[anemone]

Solve in real numbers the following equation:

$$(7x+1)^{\frac{1}{3}}+(-x^2+x+8)^{\frac{1}{3}}+(x^2-8x-1)^{\frac{1}{3}}=2.$$

 

[MarkFL]

I see by inspection that $x=0,1,9$ are solutions. (Smirk)

 

[anemone]

I see by inspection that $x=0,1,9$ are solutions. (Smirk)

Cool! But there are a total of 4 solutions to this problem...

 

[MarkFL]

Cool! But there are a total of 4 solutions to this problem...

Sweet! Then I didn't hog them all, and have left one for someone else to eyeball! (Happy)

 

[MarkFL]

I have "eyeballed" that $x=-1$ is a solution as well. (Smoking)

Time to be serious though...(Nod)

I would first arrange the equation as:

$$(7x+1)^{\frac{1}{3}}+\left(x^2-8x-1 \right)^{\frac{1}{3}}=2+\left(x^2-x-8 \right)^{\frac{1}{3}}$$

Let:

$$u=7x+1$$

$$v=x^2-8x-1$$

$$w=x^2-x-8$$

and we have:

(1) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=2+w^{\frac{1}{3}}$$

Cubing both sides of the equation, we find after simplification:

$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(2+w^{\frac{1}{3}} \right)$$

Using (1), we may write:

$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{ \frac{1}{3}} \right)$$

We may arrange this as:

$$\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)\left((uv)^{\frac{1}{3}}-2w^{\frac{1}{3}} \right)=0$$

Using the zero-factor property, this yields the cases:

i) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=0$$

$$u=-v$$

Back-substituting for $u$ and $v$, we find:

$$7x+1=-x^2+8x+1$$

$$x^2-x=0$$

$$x(x-1)=0$$

$$x=0,1$$

ii) $$(uv)^{\frac{1}{3}}-2w^{\frac{1}{3}}=0$$

$$uv=8w$$

Back substituting for $u$, $v$, and $w$, we find:

$$(7x+1)\left(x^2-8x-1 \right)=8\left(x^2-x-8 \right)$$

After simplification, we obtain:

$$(x+1)(x-1)(x-9)=0$$

$$x=-1,1,9$$

we have already ascertained that the four solutions we found are valid, hence we may state the real solutions are:

$$x=-1,0,1,9$$

 

[anemone]

Time to be serious though...(Nod)

I would first arrange the equation as:

$$(7x+1)^{\frac{1}{3}}+\left(x^2-8x-1 \right)^{\frac{1}{3}}=2+\left(x^2-x-8 \right)^{\frac{1}{3}}$$

Let:

$$u=7x+1$$

$$v=x^2-8x-1$$

$$w=x^2-x-8$$

and we have:

(1) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=2+w^{\frac{1}{3}}$$

Cubing both sides of the equation, we find after simplification:

$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(2+w^{\frac{1}{3}} \right)$$

Using (1), we may write:

$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{ \frac{1}{3}} \right)$$

We may arrange this as:

$$\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)\left((uv)^{\frac{1}{3}}-2w^{\frac{1}{3}} \right)=0$$

Using the zero-factor property, this yields the cases:

i) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=0$$

$$u=-v$$

Back-substituting for $u$ and $v$, we find:

$$7x+1=-x^2+8x+1$$

$$x^2-x=0$$

$$x(x-1)=0$$

$$x=0,1$$

ii) $$(uv)^{\frac{1}{3}}-2w^{\frac{1}{3}}=0$$

$$uv=8w$$

Back substituting for $u$, $v$, and $w$, we find:

$$(7x+1)\left(x^2-8x-1 \right)=8\left(x^2-x-8 \right)$$

After simplification, we obtain:

$$(x+1)(x-1)(x-9)=0$$

$$x=-1,1,9$$

we have already ascertained that the four solutions we found are valid, hence we may state the real solutions are:

$$x=-1,0,1,9$$

Well done, MarkFL!

But...

I have "eyeballed" that $x=-1$ is a solution as well. (Smoking)

...smoking is not a good habit and is not allowed in my thread, hehehe...:p(Smile)

 

[MarkFL]

It's an E cigarette...(Smirk)...so no worries. (Happy)