Question

X square + X + 1 is equal to zero,1 and-1​

Answer 1

GIVEN :

The quadratic equation x^2+x+1=0x2+x+1=0

TO FIND :

The value of the expression x^3x3

SOLUTION :

Given that the quadratic equation is x^2+x+1=0x2+x+1=0

Now we have to find the value of x from the given quadratic equation x^2+x+1=0x2+x+1=0

For a quadratic equation ax^2+bx+c=0ax2+bx+c=0

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac , where a and  b  are the coefficients of x^2x2 and x respectively and c is a constant.

Here a=1 ,b=1 and c=1

Substitute in the formula we get,

x=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)}x=2(1)−1±12−4(1)(1)

=\frac{-1\pm \sqrt{1-4}}{2}=2−1±1−4

=\frac{-1\pm \sqrt{-3}}{2}=2−1±−3

=\frac{-1\pm \sqrt{3i^2}}{2}=2−1±3i2 (by i^2=-1i2=−1 )

=\frac{-1\pm i\sqrt{3}}{2}=2−1±i3

∴ x=\frac{-1+i\sqrt{3}}{2}x=2−1+i3 and x=\frac{-1-i\sqrt{3}}{2}x=2−1−i3

Now we have to find x^3x3 for x=\frac{-1+i\sqrt{3}}{2}x=2−1+i3

Put x=\frac{-1+i\sqrt{3}}{2}x=2−1+i3

x^3=(\frac{-1+i\sqrt{3}}{2})^3x3=(2−1+i3)3

By using the property of exponent:

(\frac{a}{b})^m=\frac{a^m}{b^m}(ba)m=bmam

=\frac{(-1+i\sqrt{3})^3}{2^3}=23(−1+i3)3

By using the algebraic identity

(a+b)^3=a^3+3a^2b+3ab^2+b^3(a+b)3=a3+3a2b+3ab2+b3

=\frac{-1^3+3(-1)^2(i\sqrt{3})+3(-1)(i\sqrt{3})^2+(i\sqrt{3})^3}{8}=8−13+3(−1)2(i3)+3(−1)(i3)2+(i3)3

=\frac{-1+3i\sqrt{3}+9-i3\sqrt{3}}{8}=8−1+3i3+9−i33