Question

if α and β are the zeros of the quadratic polynomial x^2-2x+1 then α^3+β^3 is​

Answer 1

Answer:

$\large{\underline{\boxed{\sf \alpha^3 +\beta^3 = 2}}}$

Step-by-step explanation:

Given that ;

α and β are the zeroes of the quadratic polynomial x² - 2x + 1. On comparing the given equation with ax² + bx + c, we get -

• a = 1
• b = - 2
• c = 1

Sum of zeroes = $\bf - \dfrac{b}{a}$

⇒ α + β = $\sf \dfrac{-(-2)}{1}$

⇒ α + β = 2

Product of zeroes = $\bf \dfrac{c}{a}$

⇒ αβ = 1

$\rule{300}{2}$

We need to find, α³ + β³. So, we know that :

α³ + β³ = (α + β)³ - 3αβ(α + β)

⇒ α³ + β³ = (2)³ - 3 * 1 * 2

⇒ α³ + β³ = 8 - 6

⇒ α³ + β³ = 2

Hence, the answer is 2.

Answer 2

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{\alpha^{3}+\beta^{3}=2}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

$x^{2}-2x+1=0$

□α and β are the zeros

To find:

$\alpha^{3}+\beta^{3}=?$

${x}^{2} - 2x + 1= 0 \\ \alpha \: and \: \beta \: are \: the \: zeroes \: of \: this \: quadratic \: eqn \\ a{x}^{2} + bx + c = 0 \\ compare \: this \: eqn \: to \: given \: eqn \\ \\ a = 1 \\ b = - 2 \\ c = 1 \\ \\ \alpha + \beta = \frac{ - b}{a} = \frac{ - (- 2)}{1} = 2 \\ \\ \alpha \beta = \frac{c}{a} = \frac{1}{1} = 1 \\ \\ \alpha ^{3} + { \beta }^{2} = ( { \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) ^{2} \\ put \: the \: values \: of \: sum \: and \: product \: of \: roots \\ \alpha^{3} + \beta^{3} =( {2})^{3} - 3 \times 1(2)^{2} \\ \alpha^{3} + { \beta }^{3} = 8 - 3 \times 2 \\ { \alpha }^{ 3} + { \beta }^{ 3} = 2$

$\huge{\red{\boxed{\boxed{\green{\sf{\alpha^{3}+\beta^{3}=2}}}}}}$

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