Answer:
[tex]x=5/3\text{ or } x=1[/tex]
Step-by-step explanation:
We have the equation:
[tex]27x^6-152x^3+125=0[/tex]
We can solve this using u-substitution. Let's let u=x³. Therefore:
[tex]27(x^3)^2-152(x^3)+125=0[/tex]
Substitute all x³s for u:
[tex]27u^2-152u+125=0[/tex]
This is now in quadratic form. So, we can use the quadratic formula:
[tex]u=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Our a is 27, b is -152, and c is 125. So:
[tex]u=\frac{-(-152)\pm \sqrt{(-152)^2-4(27)(125)}}{2(27)}[/tex]
Simplify:
[tex]u=\frac{152\pm\sqrt{9604}}{54}[/tex]
Simplify:
[tex]u=\frac{152\pm 98}{54}[/tex]
Reduce. Divide everything by 2:
[tex]u=\frac{76\pm49}{27}[/tex]
Split into two cases:
[tex]u=\frac{76+49}{27}\text{ or } u=\frac{76-49}{27}[/tex]
Solve for each case:
[tex]u=\frac{125}{27}\text{ or } u=\frac{27}{27}=1[/tex]
Substitute back x³ for our u:
[tex]x^3=\frac{125}{27}\text{ or } x^3=1[/tex]
Take the cube root of each equation:
[tex]x=\sqrt[3]{\frac{125}{27}}\text{ or } x=\sqrt[3]1[/tex]
Evaluate:
[tex]x=5/3\text{ or } x=1[/tex]
And that's our two solutions.
And we're done!
