I was confused between Comp. sc., Economics and P.E., then a comp.sc. student told me that you can learn programming anytime anywhere, so why sit for exams for that as even a small silly error in writing programs can cost loosing good marks.

So now I am confused between Economics and P.E.?

What do you suggest?]]>

What about you? Computer Science?

]]>you are choosing PCM, right?along with which subject?]]>

It will start only after the results come.

what about you?]]>

We want to prove that 6(x^5 + y^5) - 15(x^4 + y^4) + 10(x^3 + y^3) - 1 = 0

Now, since cos^2 θ + sin^2 θ = 1, y=1-x

6(x^5 + y^5) = 6(x^5 + (1-x)^5) = 6 - 30 x + 60 x^2 - 60 x^3 + 30 x^4

-15(x^4 + y^4) = -15(x^4 + (1-x)^4) = -15 + 60 x - 90 x^2 + 60 x^3 - 30 x^4

10(x^3 + y^3) = 10(x^3 + (1-x)^3) = 10 - 30 x + 30 x^2

Adding up,

LHS = 6(x^5 + y^5) - 15(x^4 + y^4) + 10(x^3 + y^3) - 1 = 6 - 30 x + 60 x^2 - 60 x^3 + 30 x^4 -15 + 60 x - 90 x^2 + 60 x^3 - 30 x^4 + 10 - 30 x + 30 x^2 - 1 = 0 = RHS

##QED##

Sorry Niharka, we are still needing the binomial theorem. The above is very tedious to d by hand. I used a computer

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No one else is offering a method so here is one that will work. It's not nice and 'slick' so hopefully someone else will jump in with a simpler method.

If you expand this you'll get the powers of 10 that you need (plus some unhelpful other powers of sin and cos)

But you can write all the cos^2 terms as (1 - sin^2) instead so the unhelpful terms will all become powers of sine.

Do the same for

Now use these formulas to construct the expression

It will come to 1 without all those powers of sine, so they must all cancel out. Job done!

Bob

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