Question

If a and ß are the roots of he equation 3 x² + 6x - 2 = 0, Calculate the value a^3 - B^3​

Answer 1

Answer:

$\alpha^3-\beta^3 = \frac{28}{3}\sqrt{\frac{5}{3}}$

Step-by-step explanation:

• Given quadratic equation is: $3x^2 +6x-2=0$
• Comparing it with $ax^2+bx+c=0$, we find:

• $a = 3,\: b= 6,\: c = -2$

• It is given that: $\alpha\: and \: \beta$ are roots of the given quadratic equation. So, we find the sum and the products of the roots.

• $\implies \alpha + \beta =-\frac{b}{a}=-\frac{6}{3}=-2$

• $\& \:\alpha.\beta =\frac{c}{a}=-\frac{2}{3}$

• Since, one of the factors of $\alpha^3-\beta^3$ is $(\alpha - \beta)$ . Therefore, now we will find the value of $(\alpha - \beta),$ which can be obtained by the following formula.

• $\alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha\beta}$

• $\implies \alpha-\beta=\sqrt{(-2)^2-4\bigg(-\frac{2}{3}\bigg)}$

• $\implies \alpha-\beta=\sqrt{4+\frac{8}{3}}$

• $\implies \alpha-\beta=\sqrt{\frac{12+8}{3}}$

• $\implies \alpha-\beta=\sqrt{\frac{20}{3}}$

• $\implies \alpha-\beta=2\sqrt{\frac{5}{3}}$

• $\alpha^3-\beta^3 =(\alpha-\beta)^3+3\alpha\beta(\alpha-\beta)$

• $\implies \alpha^3-\beta^3 =\bigg(2\sqrt{\frac{5}{3}}\bigg)^3+3\bigg(-\frac{2}{3}\bigg)\bigg(2\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 =8\bigg(\sqrt{\frac{5}{3}}\bigg)^3-\bigg(\frac{6}{3}\bigg)\bigg(2\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 =8\times \frac{5}{3}\bigg(\sqrt{\frac{5}{3}}\bigg)-4\bigg(\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 = \frac{40}{3}\bigg(\sqrt{\frac{5}{3}}\bigg)-4\bigg(\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 = \bigg(\frac{40}{3}-4\bigg)\bigg(\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 = \bigg(\frac{40-12}{3}\bigg)\bigg(\sqrt{\frac{5}{3}}\bigg)$

• $\implies \alpha^3-\beta^3 = \frac{28}{3}\sqrt{\frac{5}{3}}$