Question

if X square + X + 1 is equal to zero then find the value of x cube​

Answer 1

GIVEN :

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

TO FIND :

The value of the expression $x^3$

SOLUTION :

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

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

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

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

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

Substitute in the formula we get,

$x=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)}$

$=\frac{-1\pm \sqrt{1-4}}{2}$

$=\frac{-1\pm \sqrt{-3}}{2}$

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

$=\frac{-1\pm i\sqrt{3}}{2}$

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

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

Put $x=\frac{-1+i\sqrt{3}}{2}$

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

By using the property of exponent:

$(\frac{a}{b})^m=\frac{a^m}{b^m}$

$=\frac{(-1+i\sqrt{3})^3}{2^3}$

By using the algebraic identity

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

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

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

$=\frac{-1+9}{8}$

$=\frac{8}{8}$

$=1$

∴  $x^3=1$ for $x=\frac{-1+i\sqrt{3}}{2}$

Now we have to find $x^3$ for $x=\frac{-1-i\sqrt{3}}{2}$

Put $x=\frac{-1-i\sqrt{3}}{2}$

$x^3=(\frac{-1-i\sqrt{3}}{2})^3$

$=\frac{(-1-i\sqrt{3})^3}{2^3}$

By using the algebraic identity

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

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

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

$=\frac{-1+9}{8}$

$=\frac{8}{8}$

$=1$

∴  $x^3=1$ for $x=\frac{-1-i\sqrt{3}}{2}$

∴ the value of $x^3=1$ for $x=\frac{-1+i\sqrt{3}}{2}$ and $x^3=1$ for $x=\frac{-1-i\sqrt{3}}{2}$

Answer 2

Answer:

GIVEN :

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

2

+x+1=0

TO FIND :

The value of the expression x^3x

3

SOLUTION :

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

2

+x+1=0

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

2

+x+1=0

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

2

+bx+c=0

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}x=

2a

−b±

b

2

−4ac

, where a and b are the coefficients of x^2x

2

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±

1

2

−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±

3i

2

(by i^2=-1i

2

=−1 )

=\frac{-1\pm i\sqrt{3}}{2}=

2

−1±i

3

∴ x=\frac{-1+i\sqrt{3}}{2}x=

2

−1+i

3

and x=\frac{-1-i\sqrt{3}}{2}x=

2

−1−i

3

Now we have to find x^3x

3

for x=\frac{-1+i\sqrt{3}}{2}x=

2

−1+i

3

Put x=\frac{-1+i\sqrt{3}}{2}x=

2

−1+i

3

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

3

=(

2

−1+i

3

)

3

By using the property of exponent:

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

b

a

)

m

=

b

m

a

m

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

2

3

(−1+i

3

)

3

By using the algebraic identity

(a+b)^3=a^3+3a^2b+3ab^2+b^3(a+b)

3

=a

3

+3a

2

b+3ab

2

+b

3

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

8

−1

3

+3(−1)

2

(i

3

)+3(−1)(i

3

)

2

+(i

3

)

3

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

8

−1+3i

3

+9−i3

3

=\frac{-1+9}{8}=

8

−1+9

=\frac{8}{8}=

8

8

=1=1

∴ x^3=1x

3

=1 for x=\frac{-1+i\sqrt{3}}{2}x=

2

−1+i

3

Now we have to find x^3x

3

for x=\frac{-1-i\sqrt{3}}{2}x=

2

−1−i

3

Put x=\frac{-1-i\sqrt{3}}{2}x=

2

−1−i

3

x^3=(\frac{-1-i\sqrt{3}}{2})^3x

3

=(

2

−1−i

3

)

3

=\frac{(-1-i\sqrt{3})^3}{2^3}=

2

3

(−1−i

3

)

3

By using the algebraic identity

(a-b)^3=a^3-3a^2b+3ab^2-b^3(a−b)

3

=a

3

−3a

2

b+3ab

2

−b

3

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

8

−1

3

+3(−1)

2

(−i

3

)+3(−1)(−i

3

)

2

+(−i

3

)

3

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

8

−1−3i

3

+9+i3

3

=\frac{-1+9}{8}=

8

−1+9

=\frac{8}{8}=

8

8

=1=1

∴ x^3=1x

3

=1 for x=\frac{-1-i\sqrt{3}}{2}x=

2

−1−i

3

∴ the value of x^3=1x

3

=1 for x=\frac{-1+i\sqrt{3}}{2}x=

2

−1+i

3

and x^3=1x

3

=1 for x=\frac{-1-i\sqrt{3}}{2}x=

2

−1−i

3