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
`(-b+ √(b^2-4ac) )/(2a)` and subsitute the value of
`a, b, c`


`p_1= (-(-98)+ √((-98)^2-4(1)(-3375)) )/(2) =125`


`p_2= (-(-98)- √((-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