IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

(1) how can i solve the expression

(x-a)/b+(x-b)/a=b/(x-a)+a/(x-b)

(2) if x³-x²-x-1 has a root a,b,c

than prove that (a¹⁹⁹²-b¹⁹⁹²)/a-b+ (b¹⁹⁹²-c¹⁹⁹²)/b-c+ (c¹⁹⁹²-a¹⁹⁹²)/c-a

has integer.

shubhayan_c

(ax - a^2 + bx - b^2)/ ab = (ax - a^2 + bx - b^2)/(x-a)(x-b)

1/ab = 1/(x-a)(x-b)

x^2 - (a +b)x + ab = ab

x^2 = (a+b)x

x = (a+b)

Greatdreams

1.)

(ax - a^2 + bx - b^2)/ ab - (ax - a^2 + bx - b^2)/(x-a)(x-b) = 0

So (ax - a^2 + bx - b^2)/ ab - (ax - a^2 + bx - b^2)/(x-a)(x-b) = 0

(ax - a^2 + bx - b^2)[1/ab - 1/(x-a)(x-b)] = 0

So (ax - a^2 + bx - b^2) (x^2 - ax - bx) = 0

thus [x(a+b) - a^2 - b^2] [ x^2 - x(a+b)] = 0

So we have

x(a+b) - a^2 - b^2 = 0 ; x = a^2 + b^2/a+b

and x^2 - x(a+b) = 0

So x[x - (a+b)] = 0

So x = 0 and x = a+ b

So the values are x = 0 , a+b , a^2 + b^2/a+b

as for the 2nd question i think some data is missing

it should be x3 - x2 - x +1 = 0

hsbhatt

x-a/a + x-b/b = b/x-a + a/x-b

=> x-a/a - a/x-b = b/x-a - x-b/a

or (x-a)(x-b)-ab/b(x-b) = ab-(x-a)(x-b)/a(x-a)

Hence, either (x-a)(x-b) = ab or b(x-b) = a(x-a)

Solving we get x = 0 or x = a+b or x = a-b

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