Question

Here's Your "SwaggerGabru"

Harsh Pratap Singh :)

QUESTION :-
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Solve the problem in attachment!!!

(For me this is too important question if you Don't know plz don't answer and if you want points then Don't worry I am continously posting questions.)

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Note:-
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Answer should be easy to understand.

(You guys just Spammed my every question whatever questions I asked in last hours are all wrong except few.) ​​

Answer 1

Let P(x)=a0+a1x+⋯+anxn. For every x∈R the triple (a,b,c)=(6x,3x,−2x) satisfies the condition ab+bc+ca=0. Then the condition on P gives us P(3x)+P(5x)+P(−8x)=2P(7x) for all x, implying that for all i=0,1,2,…,n the following equality holds:

(3i+5i+(−8)i−2⋅7i)ai=0.

Suppose that ai≠0. Then K(i)=3i+5i+(−8)i−2⋅7i=0. But K(i) is negative for i odd and positive for i=0 or i≥6 even. Only for i=2 and i=4 do we have K(i)=0. It follows that P(x)=a2x2+a4x4 for some real numbers a2,a4. It is easily verified that all such P(x) satisfy the required condition.

Answer 2

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