Question

If a+b+c=0, prove that a^2/bx+b^2/cx+c^2/ax=3.​

Answer 1

The equation x^2+ax+b = 0 has two roots

x1 = [-a +(a^2–4b)^0.5]/2 …(1) and

x2 = [-a -(a^2–4b)^0.5]/2 …(2)

The other equation x^2+bx+a = 0 has two roots

x1 = [-b +(b^2–4a)^0.5]/2 …(3) and

x2 = [-b -(b^2–4a)^0.5]/2 …(4)

Since the roots are the same for both equations, equate x1 from (1) to x1 from (3). Or

[-a +(a^2–4b)^0.5]/2 = [-b +(b^2–4a)^0.5]/2, or multiplying by 2 on both sides

[-a +(a^2–4b)^0.5] = [-b +(b^2–4a)^0.5], or

(a^2–4b)^0.5 - (b^2–4a)^0.5 = a - b. Squaring both sides

(a^2–4b) + (b^2–4a) -2(a^2–4b)^0.5*(b^2–4a)^0.5 = a^2 - 2ab + b^2, or

a^2+b^2 -(4a+4b) -a^2+2ab - b^2 = 2(a^2–4b)^0.5*(b^2–4a)^0.5, or

2ab-4a-4b = 2(a^2–4b)^0.5*(b^2–4a)^0.5, or

ab -2a-2b = (a^2–4b)^0.5*(b^2–4a)^0.5. Again square both sides.

4a^2+4b^2+a^2b^2–4a^2b-4ab^2+8ab = (a^2–4b)*(b^2–4a) = a^2b^2–4b^3–4a^3+16ab, or

4a^3+4b^3+4a^2+4b^2+a^2b^2–4a^2b-4ab^2+8ab -a^2b^2–16ab = 0, or

4a^3+4b^3+4a^2+4b^2–4a^2b-4ab^2-8ab = 0, or dividing by 4

a^3+b^3+a^2+b^2–a^2b-ab^2–2ab = 0

(a^3+b^3)+(a^2–2ab+b^2)–(a^2b+ab^2) = 0, or

(a+b)(a^2-ab+b^2) + (a-b)^2 -ab(a+b) = 0, or

(a+b)(a^2-ab+b^2-ab) + (a-b)^2 =0, or

(a+b)(a^2–2ab+b^2) +(a-b)^2 =0, or

(a+b)(a-b)^2 +(a-b)^2 =0, or

(a+b+1)(a-b)^2=0.

Hence a+b+1 = 0. Proved.