x^6-9x^3+8=0 plz solve the equation
Answers
Answer:
x⁶-9x³+8 = (x-2) (x²+2x+4) (x-1) (x²+x+1)
So
x=2 and
x=1
are the real solutions
Step-by-step explanation:
This polynomial is quadratic in
x³ ,factoring into two cubic factors, which then factor a little further..
0=x⁶−9x³+8
= (x³)²- (8+1) (x³) +(8×1)
=(x³-8) (x³-1)
= (x³_2³) (x³-1³)
=(x-2) (x²+2x+4) (x-1) (x²+x+1)
Answer:
Step-by-step explanation:
x6 - 9x3 + 8 = 0 [Given]
Consider u = x3
So the given equation becomes a quadratic.
u2 - 9u + 8 = 0
By splitting the middle term, we solve the quadratic equation.
u2 - 1u - 8u + 8 = 0
Taking u as common in the first two terms and 8 as common in the next two terms
u (u - 1) - 8 (u - 1) = 0
(u - 8) (u - 1) = 0
So we get
u - 8 = 0 or u - 1 = 0
u = 8 or u = 1
Substituting the value of u
x3 = 1
x3 = 8
So we get x = 1, 2.
Therefore, the solutions of the equation are x = 1, 2.
The solutions of the equation x6 - 9x3 + 8 = 0 using u substitution are x = 1, 2.
Poojaa!