Question

$x2 - 5x + 6 = 0 \\ then \: find \\\alpha 3 - \beta 3$
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Answer 1

x²-5x+6 =0

=>x² -(3x+2x)+6 = 0

=> x² - 3x-2x + 6 = 0

=>(x²-3x)+(-2x+6) = 0

=>x(x-3) +-2(x-3) = 0

=>(x-2)+(x-3) = 0

~> x- 2 = 0 so, x = 2

~> x-3 = 0 so, x = 3

Here,

$\alpha = 2 \: \: and \beta = 3$

${ \alpha }^{3} - { \beta }^{3} = 2 {}^{3} - {3}^{3} = 8 - 27 =-19 (or 19)$

Cheers!

Answer 2

Correct Question :-

x² - 5x + 6 = 0 then find $\sf \alpha^3 - \beta^3$

Solution :-

x² - 5x + 6 = 0

Split the middle term

⇒ x² - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 3)(x - 2) = 0

⇒ x - 3 = 0 , x - 2 = 0

⇒ x = 3, x = 2

Here,

$\sf (i) \: If \: \alpha = 3 \: \: \beta = 2$

$\sf \alpha^3 - \beta^3$

$\sf = {(3)}^{3} - {(2)}^{3}$

$\sf = 27 - 8$

$\sf =19$

$\sf (ii) \: If \: \alpha = 2 \: \: \beta = 3$

$\sf \alpha^3 - \beta^3$

$\sf = {(2)}^{3} - {(3)}^{3}$

$\sf = 8 - 27$

$\sf = - 19$

$\sf \alpha^3 - \beta^3 = - 19$

$\bf \alpha^3 - \beta^3 =19 \: or \: - 19$