Question

Find the zeroes of the polynomial, f(x) = x^3-1​

Answer 1

Answer:

x=1,-1+√-3/2,-1-√-3/2

f(x)=x^3-1

we know that:a^3-b^3=(a-b)(a^2+b^2+ab)

so,f(x)becomes (x-1)(x^2+x+1)

then,x-1=0 and by solving (x-1)(x^2+x+1) by quadratic formula,we get x=-1+√1-4/2

x=-1+-√-3/2

so the roots of the given polynomial are:

x=1,-1+√-3/2and -1-√-3/2