please meritnation experts solve these questions for me as i have a really important test tomorrow i really need your help if a polynomial p(x)=Ax^3+Bx^2+Cx+D vanishes at x= a-d,a+d,a,then prove that [a^2+(D/aA)]0.Here A,B,C,D are some constants. If alpha and beta are zeroes of x^2-3x+1 then find the value of (alpha ^2014 +alpha^2016+ beta^2014+ beta^2014)/(alpha^2015+ beta^2015) solve for x and y (x/a^2) + (y/b^2) = 2/ab and (x/2b) - (y/2a) = (a^3-b^3) / (2a^2b^2) Share with your friends Share 0 Anuradha Sharma answered this 3) We have,xa2 + yb2 = 2ab ...............1x2b - y2a = a3 - b32 a2b2 ...............2Multiplying 1 by 12b and 2 by 1a2 we get,x2a2b + y2b3 = 1ab2 ..............3x2 a2b - y2 a3 = a3 - b32 a4b2 ...............4Subtracting 4 from 3 we get, y2 b3 + y2 a3 = 1ab2 - a3 - b32 a4b2⇒ y 12b3 + 12a3 = 1ab2 1 - a3 - b32a3⇒ y a3 + b32 a3b3 = 1ab2 2a3- a3 - b32a3⇒ y a3 + b32 a3b3 = 1ab2 2a3 - a3 + b32a3⇒ y a3 + b32 a3b3 =1ab2 a3 + b32a3 ⇒ y = 1ab2 a3 + b32a3 × 2a3 b3a3 + b3⇒ y = 1ab2 × 12a3 × 2 a3b3⇒ y = 2 a3 b32 a4 b2⇒ y = baPutting the value of y in 1, we get,xa2 + bab2 = 2ab⇒ xa2 + bab2 = 2ab⇒ xa2 + 1ab = 2ab⇒ xa2 = 2ab - 1ab⇒ xa2 = 2 - 1ab⇒ xa2 = 1ab⇒ x = a2ab ⇒ x = abHence, x = ab and y = ba 2) Given α and β are the roots of the equation x2-3x+1=0 so, they will satisfy the equation , α2-3α+1=0 or α2+1=3α...(1)β2-3β+1=0 or β2+1=3β...(2)Now we have to find , α2014+α2016+β2014+β2016α2015+β2015 =α20141+α2+β20141+β2α2015+β2015Now putting value from equation (1) and (2) we get, =3α2015+3β2015α2015+β2015=3α2015+β2015α2015+β2015=3 0 View Full Answer