If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc"

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Mainak Das

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u can do this problem with the help of SHIDHARACHARYA formula which is ax2=bx+c=0
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apply shri dharacharai identity....when both roots are equal B^2-4AC=0 ..........(1) HERE B=-2(a^2-bc), A=c^2-ab, C=b^2-ac now apllying the conditions,and put all these values in eqn (1) we get 4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0 a^4+b^3+c^3-3a^2bc=0 by taking common a from above eqn we get what...
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apply shri dharacharai identity....when both roots are equal B^2-4AC=0 ..........(1) HERE B=-2(a^2-bc), A=c^2-ab, C=b^2-ac now apllying the conditions,and put all these values in eqn (1) we get 4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0 a^4+b^3+c^3-3a^2bc=0 by taking common a from above eqn we get what we have to prove it a=0 or a^3+b^3+c^3=3abc read less
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Eco, Maths and Stats Tutor

Basically, remember for a quadratic equation Ax^2 + Bx +C=0, the two roots are equal only if B^2 - 4AC=0. In the given equation, A=(c^2 - ab), B = -2(a^2-bc) and C = b^2 - ac. Substitute these values in the relation B^2 =4AC and simplify to get the desired result.
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(a^2-bc)^2=(c^2-ab)(b^2-ac) a(a^3-2abc)=a(-b^3-c^3+abc) a=0(or)a^3+b^3+c^3=3abc
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Good basic mathematics, Basic electronics, VLSI, Basic technical knowledge

Both roots are equal=> D=0 => b2-4ac=0=> 4 (a2-bc)2-4 (c2-ab)(b2-ac)=0 =>a4-2a2bc+b2c2-b2c2+c3a+b3a-a2bc=0 => a (a3+b3+c3-2abc-abc)=0; for the equation to be 0, either a=0 or a3+b3+c3-3abc=0 ; so a=0 or a3+b3+c3=3abc
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