q1 if the roots of the equation (c^-ab)x^-2(a^-bc)x+b^-ac=0 are equal prove that either a=0 or a cube+b cube+c cube=3abc? q2 a polygon of n sides has n(n-3)/2 diagonals how many sides has a polygon with 54 diagonals? q3 if sin alpha and cos alpha are thr roots of the equation ax^+bx+c=0 then provethat a^+2ac=b^? q4 if the roots of the equation (a^+b^)x^-2(ac+bd)x+(c^+d^)=0 are equal then prove that a/b=c/d?

Please ask one question at a time. The answer to your first two questions is given below.

 

1) We know that for real and equal roots D=0

so, b²-4ac = 0

{-2(a²-bc)}² - 4 (c²-ab) (b²-ac) = 0

4(a²-bc)² - 4(c²-ab)(b²-ac)=0

4( a⁴+b²c²-2a²bc) = 4(b²c²-ac³-ab³+a²bc)

a⁴+ac³+ab³ = 2a²bc+a²bc

a (a³+c³+b³) = 3a²bc

Hence a = 0 and a³+b³+c³ = 3abc

2) a polygon of n sides has n(n-3)/2 diagonals

For the required polygon, number of diagonals = 54

So, n(n-3)/2 = 54

n² - 3n - 108 = 0

n² - 12n + 9n - 108 = 0

(n - 12) (n + 9) = 0

Since, n cannot be negative, n = 12

So, the number of sides of the required polygon is 12.