Question

Please solve this friends ​

Answer 1

Answer:

(1/alpha³) + (1/beta³) = (3ab/c²) - (b³/c³)

Step-by-step explanation:

We have,

ax² + bx + c = 0

and A (alpha) and B (beta) are the zeroes.

Now,

We know that,

Sum of Zeroes = -b/a

A + B = -b/a ---- 1

Similarly,

Product of zeroes = c/a

AB = c/a ---- 2

Let's cube it so that we will get (1/A³) + (1/B³)

Now,

(x + y)³ = x³ + 3xy(x + y) + y³

Then,

[(1/A) + (1/B)]³ = (1/A)³ + 3(1/A)(1/B)[(1/A) + (1/B)] + (1/B)³

Now,

(1/A) + (1/B)

Taking LCM = AB,

(B/AB) + (A/AB)

= [(A + B)/AB]

Now,

[(A + B)/AB]

From eq.1 and eq.2,

= [(-b/a)/(c/a)]

= (-b/a) × (a/c)

= (-b/c)

Hence,

[(A + B)/AB)]³ = (1/A³) + 3(1/AB)[(A + B)/AB] + (1/B³)

(-b/c)³ = (1/A³) + 3[1/(c/a)](-b/c) + (1/B³)

(-b³/c³) = (1/A³) + (1/B³) + 3(a/c)(-b/c)

(-b³/c³) = (1/A³) + (1/B³) + 3(-ab/c²)

(-b³/c³) = (1/A³) + (1/B³) - (3ab/c²)

(1/A³) + (1/B³) = (3ab/c²) - (b³/c³)

Hence,

(1/alpha³) + (1/beta³) = (3ab/c²) - (b³/c³)

Hope it helped and believing you understood it........All the best