Rearrange [tex]x^6-98x^3=3375[/tex] to give
[tex]x^6-98x^3-3375=0[/tex]
Let [tex]x^3[/tex] be a variable 'p' and so we can write [tex]x^6[/tex] as [tex]x^6=(x^3)(x^3)=(p)(p)= p^2[/tex]
Rewrite the equation in terms of 'p'
[tex]p^2-98p-3375 = 0 [/tex]
where [tex]a=1, b=-98, c=-3375[/tex]
using the quadratic formula [tex] \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex] and subsitute the value of [tex]a, b, c[/tex]
[tex]p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125[/tex]
[tex]p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27[/tex]
There are two value of p; 125 and -27
Now we find the value of x
Earlier we substitute [tex]x^3 [/tex] for [tex]p[/tex] and [tex]x^6[/tex] for [tex]p^2[/tex]
When [tex]p=125[/tex], [tex]x= \sqrt[3]{125}=5 [/tex]
When [tex]p = -27, x= \sqrt[3]{-27}=-3 [/tex]
So the final answer is the two values of x;
x = 5 OR x = -3
