Solve An Equation

anemone

MHB POTW Director
Staff member
Solve in real numbers the following equation:

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

MarkFL

Staff member
I see by inspection that $x=0,1,9$ are solutions.

anemone

MHB POTW Director
Staff member
I see by inspection that $x=0,1,9$ are solutions.
Cool! But there are a total of 4 solutions to this problem...

MarkFL

Staff member
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!

MarkFL

Staff member
I have "eyeballed" that $x=-1$ is a solution as well.

Time to be serious though...

I would first arrange the equation as:

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

Let:

$$\displaystyle u=7x+1$$

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

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

and we have:

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

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

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

Using (1), we may write:

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

We may arrange this as:

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

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

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

$$\displaystyle u=-v$$

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

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

$$\displaystyle x^2-x=0$$

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

$$\displaystyle x=0,1$$

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

$$\displaystyle uv=8w$$

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

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

After simplification, we obtain:

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

$$\displaystyle x=-1,1,9$$

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

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

anemone

MHB POTW Director
Staff member
Time to be serious though...

I would first arrange the equation as:

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

Let:

$$\displaystyle u=7x+1$$

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

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

and we have:

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

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

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

Using (1), we may write:

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

We may arrange this as:

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

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

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

$$\displaystyle u=-v$$

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

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

$$\displaystyle x^2-x=0$$

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

$$\displaystyle x=0,1$$

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

$$\displaystyle uv=8w$$

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

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

After simplification, we obtain:

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

$$\displaystyle x=-1,1,9$$

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

$$\displaystyle x=-1,0,1,9$$
Well done, MarkFL!

But...

I have "eyeballed" that $x=-1$ is a solution as well.
...smoking is not a good habit and is not allowed in my thread, hehehe...