Question

if alpha and beta are the zeros of the polynomial x^2+x+1 find the value of alpha^3 + beta^3​

Answer 1

Given:

• p(x) = x² + x + 1

• α and β are the zeroes of this polynomial.

To find:

• α³ + β³

Solution:

We know that;

• Sum of zeroes (α + β) = -b/a

• Product of zeroes (αβ) = c/a

Where:

• a - Coefficient of x².

• b - Coefficient of x.

• c - Constant.

Therefore;

⇒ Sum of zeroes (α + β) = -b/a

⇒ Sum of zeroes (α + β) = -1/1

⇒ Sum of zeroes (α + β) = -1

⇒ Product of zeroes (αβ) = c/a

⇒ Product of zeroes (αβ) = 1/1

⇒ Product of zeroes (αβ) = 1

We know the identity (a + b)³ = a³ + b³ + 3ab(a + b)

∴ a³ + b³ = (a + b)³ - 3ab(a + b)

Now, Let's find the value of α³ + β³.

⇒ α³ + β³

⇒ (α + β)³ - 3αβ(α + β)

⇒ (-1)³ - 3(1)(-1)

⇒ -1 + 3

⇒ 2

∴ α³ + β³ = 2

Answer 2

Solution :

$\bigstar$ We have quadratic polynomial p(x) = x² + x + 1

As we know that given polynomial compared with ax² + bx + c;

• a = 1
• b = 1
• c = 1

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg(\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } \bigg)}\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{1}{1} }\\\\\\\mapsto\bf{\alpha +\beta =-1}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg(\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } \bigg)}\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{1}{1} }\\\\\\\mapsto\bf{\alpha \times \beta =1}$

So;

$\mapsto\tt{\alpha ^{3} +\beta ^{3} }=(\alpha +\beta )^{3} -3\alpha \beta(\alpha +\beta )}\\\\\mapsto\tt{\alpha ^{3} +\beta ^{3}=(-1)^{3} -3\times 1(-1)}\\\\\mapsto\tt{\alpha ^{3} +\beta ^{3}=-1-3\times (-1)}\\\\\mapsto\tt{\alpha ^{3} +\beta ^{3}=-1+3}\\\\\mapsto\bf{\alpha ^{3} +\beta ^{3}=2}$

Thus;

The value of α³ + β³ will be 2 .