Question

pls tell this ans with full solutuon you will solve i will mark you brainliest​

Answer 1

Solution

If Ax2 + Bx + C = 0 has equal roots, then B2 = 4AC. Using this, we have

(2(b2 - ac))2 = 4×(a2 - bc)(c2 - ab)

⇨ (b2 - ac)2 = (a2 - bc)(c2 - ab)

⇨ b4 + a2c2 - 2ab2c = a2c2 - a3b - bc3 + ab2c

⇨ b4 - 2ab2c = - a3b - bc3 + ab2c

⇨ b4 + a3b + bc3 = 2ab2c + ab2c

⇨ b (a3 + b3 + c3) = b(3abc)

⇨ a3 + b3 + c3 = 3abc

The correct option is B.

Answer 2

Since the equation has equal roots,

D = b²-4ac = 0

In this equation,

a = a²-bc

b = 2(b²-ac)

c = c²-ab

Putting values of a, b, c in D

b²-4ac = 0

[2(b²-ac)]² - 4*(a²-bc)*(c²-ab) = 0

4(b⁴+a²c²-2ab²c) -4(a²c² - a³b - bc³ + ab²c) = 0

4(b⁴+a²c²-2ab²c) = 4(a²c² - a³b - bc³ + ab²c)

b⁴+a²c²-2ab²c = a²c² - a³b - bc³ + ab²c

b⁴ - 2ab²c = -a³b - bc³ + ab²c       (a²c² cancelled from both sides)

b⁴ = ab²c(2+1) - a³b - bc³          (2ab²c taken to RHS and added to ab²c)

b⁴ = 3ab²c - a³b - bc³

b³ = 3abc - a³ - c³                       (b cancelled from all the terms)

a³+b³+c³ = 3abc

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