To solve the equation [tex]\(4 x^{-\frac{7}{3}} + 3 x^{-\frac{4}{3}} - x^{-\frac{1}{3}} = 0\)[/tex], we can proceed as follows:
1. Variable Substitution: Let's make a substitution to simplify the exponents. Set [tex]\( y = x^{-\frac{1}{3}} \)[/tex]. Then we have:
- [tex]\(x^{-1/3} = y\)[/tex]
- [tex]\(x^{-4/3} = (x^{-1/3})^4 = y^4\)[/tex]
- [tex]\(x^{-7/3} = (x^{-1/3})^7 = y^7\)[/tex]
Substituting these into the original equation, we get:
[tex]\[
4y^7 + 3y^4 - y = 0
\][/tex]
2. Factoring: Notice that [tex]\( y = 0 \)[/tex] is not a valid solution because it corresponds to [tex]\( x \to \infty \)[/tex], which is not defined in this context. Therefore, we can factor out [tex]\(y\)[/tex]:
[tex]\[
y(4y^6 + 3y^3 - 1) = 0
\][/tex]
This gives us two factors which can be separately set to zero:
[tex]\[
y = 0 \quad (\text{not a valid solution, as mentioned})
\][/tex]
[tex]\[
4y^6 + 3y^3 - 1 = 0
\][/tex]
3. Solving the Polynomial: Now we need to solve the polynomial equation [tex]\(4y^6 + 3y^3 - 1 = 0\)[/tex]. This is highly non-trivial, but let us explore possible roots:
- By inspection and numerical methods (given the nature of the polynomial), we can obtain the roots for [tex]\(y\)[/tex].
4. Finding the Solutions: The roots of [tex]\(4y^6 + 3y^3 - 1 = 0\)[/tex] are found to be:
[tex]\[
y = -1 \quad (\text{real root})
\][/tex]
[tex]\[
y = 4 \quad (\text{real root})
\][/tex]
[tex]\[
y \approx -1 \pm 0i \quad (\text{considered as a single root at } y = -1)
\][/tex]
[tex]\[
y \approx 4 \pm 0i \quad (\text{considered as a single root at } y = 4)
\][/tex]
5. Back Substitution: Convert back from [tex]\(y\)[/tex] to [tex]\(x\)[/tex]:
[tex]\[
y = x^{-1/3} \implies x = y^{-3}
\][/tex]
Therefore, for [tex]\(y = -1\)[/tex]:
[tex]\[
x = (-1)^{-3} = -1
\][/tex]
For [tex]\(y = 4\)[/tex]:
[tex]\[
x = 4^{-3} = 1/64
\][/tex]
Thus, the solutions to the original equation [tex]\(4 x^{-\frac{7}{3}} + 3 x^{-\frac{4}{3}} - x^{-\frac{1}{3}} = 0\)[/tex] are:
[tex]\[
x = -1 \quad \text{or} \quad x = 1/64
\][/tex]
