MarkFL said:I see by inspection that $x=0,1,9$ are solutions. (Smirk)
anemone said:Cool! But there are a total of 4 solutions to this problem...
MarkFL said:Time to be serious though...(Nod)
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:
\(\displaystyle x=-1,0,1,9\)
MarkFL said:I have "eyeballed" that $x=-1$ is a solution as well. (Smoking)
The equation to be solved is (7x+1)^(1/3)+(-x^2+x+8)^(1/3)+(x^2-8x-1)^(1/3)=2.
The purpose of solving this equation is to find the value(s) of x that make the equation true.
The degree of this equation is 3, as indicated by the cube roots in each term.
Yes, this equation can be solved algebraically by isolating the variable and raising both sides of the equation to the appropriate power.
Yes, there are restrictions on the value of x for this equation. The radicands (expressions within the radical signs) must be non-negative, so x cannot be equal to -1 or -8. Additionally, the denominator of the fraction must be non-zero, so x cannot be equal to 0.