- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

- Admin
- #2

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,909

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

- Admin
- #4

Sweet! Then I didn't hog them all, and have left one for someone else to eyeball!Cool! But there are a total of 4 solutions to this problem...

- Admin
- #5

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:

\(\displaystyle x=-1,0,1,9\)

- Thread starter
- Admin
- #6

- Feb 14, 2012

- 3,909

Well done,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:

\(\displaystyle x=-1,0,1,9\)

But...

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

- Admin
- #7