Question

Solve the equation after making an appropriate substitution. X^6-98x^3=3375

Answer 1

Rearrange $x^6-98x^3=3375$ to give

$x^6-98x^3-3375=0$

Let $x^3$ be a variable 'p' and so we can write $x^6$ as $x^6=(x^3)(x^3)=(p)(p)= p^2$

Rewrite the equation in terms of 'p'

$p^2-98p-3375 = 0$

where $a=1, b=-98, c=-3375$

using the quadratic formula $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ and subsitute the value of $a, b, c$

$p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125$

$p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27$

There are two value of p; 125 and -27 

Now we find the value of x

Earlier we substitute $x^3$ for $p$ and $x^6$ for $p^2$

When $p=125$, $x= \sqrt[3]{125}=5$
When $p = -27, x= \sqrt[3]{-27}=-3$

So the final answer is the two values of x; 

x = 5 OR x = -3