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What a brilliant question; R different from Q is a great hint. R-Q different from zero --> denominator of a rational number cannot be zero + I like difference of squares = subtract equations and simplify. Had I noticed that small nuance when I was solving, I would not have gotten it wrong. Alas, It will come in handy next time.
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learn the rules of the game, then play better than anyone else.

Bunuel, could you explain the mental steps you took to get to the different steps in your solution? I would not have even known to get to some of those steps.

The clue helped, but I still couldn't bridge how to eliminate the other variables with what I had. The logic which I'm to substitute or combine is kind of ambiguous now, and this problem exploited that weakness. I found the article below, but any body have other guidance on how to step through expressions in a logical manner?

The clue helped, but I still couldn't bridge how to eliminate the other variables with what I had. The logic which I'm to substitute or combine is kind of ambiguous now, and this problem exploited that weakness. I found the article below, but any body have other guidance on how to step through expressions in a logical manner?

This is quite painful topic for me too. And I still struggle with it occasionally but I'll try to help to you with this one:

firstly we have three equations: \(P^2 - QR = 10\) \(Q^2 + PR = 10\) \(R^2 + PQ = 10\)

We see that each pair of them have three pair of the same values. Let's take for example another pair (not which Bunuel describe in his solution) The method will be same:

\(P^2 - QR = 10\) \(Q^2 + PR = 10\)

our goal is to get rid of squares and received clear values we can make it by using formula \(a^2-b^2=(a-b)(a+b)\)

here is a lot about algebra and exactly about these Algebraic Identities math-algebra-101576.html (at the end of the article in subsection Algebraic Identities)

So we need to make from these two equations \(P^2 - QR = 10\) \(Q^2 + PR = 10\)

one equation without exponents and we can see that by subtraction we will receive \(Q^2-P^2\) so we can try to check this idea: \(Q^2 + PR - P^2 + QR = 0\) \(Q^2 - P^2 + PR + QR = 0\) \((Q - P)(Q + P) + PR + QR = 0\)

Now we need to receive \(Q - P\) or \(Q + P\) for cancellation this equation from this part \(PR + QR\) we can receive \(R(Q+P)\) So we have: \((Q - P)(Q + P) - R(Q+P) = 0\) --> \((Q - P)(Q + P) = -R(Q+P)\) --> cancel \((Q + P)\) from both sides --> \((Q - P) = -R\) or \(P = R + Q\)

And now we should see the way to eliminate these obnoxious parts \(PR\) and \(PQ\) Again we should take a look on all three equations and try to see what we can eliminate to receive needed result. \(P^2 - QR = 10\) \(Q^2 + PR = 10\) \(R^2 + PQ = 10\)

We have part \(Q+R = P\) so ideally will be to find parts in which we have the same disguised expression These two equations have it \(Q^2 + PR = 10\) \(R^2 + PQ = 10\)

Let's sum them: \(Q^2 + R^2 + PR + PQ = 20\) --> undisguise \(P + Q\) --> \(Q^2 + R^2 + P(R + Q) = 20\) And now we have expression from our question \(Q^2 + R^2 + P^2 = 20\)

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For solving such tasks you should know Algebraic identites by heart and do a lot of such tasks with thinking about why author of solution pick exactly these values to cancel or to transform in another view

I think after 50-70 solved and dissected tasks you will obtain "feeling" of "where is hide identities which will help to solve the task" and this process even can became boring for 600 lvl tasks Good luck with your preparation.
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

Hi sindhubharadwaj and shasadou.., i'll try and explain in a shorter method, but it wou;ld also mean playing with the Equations..

Quote:

If P2−QR=10, Q2+PR=10, R2+PQ=10, and R≠Q, what is the value of P2+Q2+R2?

A. 10 B. 15 C. 20 D. 25 E. 30

lets pick two equations containing R^2 and Q^2 terms, as it is given R≠Q.. Q^2+PR=10, R^2+PQ=10.. subtract two .. Q^2+PR-( R^2+PQ)=10-10.. Q^2-R^2+PR-PQ=0.. (Q-R)(Q+R)-P(Q-R)=0.. (Q+R-P)(Q-R)=0 As it is given R≠Q, so (Q+R-P)=0.. so P=R+Q..(i)

Add all three equations.. P^2−QR+ Q^2+PR+R^2+PQ=30, P^2+ Q^2+R^2+PR+PQ−QR=30, (P^2+ Q^2+R^2)+PR+PQ−QR=30, (P^2+ Q^2+R^2)+PR+PQ−QR=30, (P^2+ Q^2+R^2)+P(R+Q)−QR=30,..(ii)

from (i), we get P=R+Q.. substitute in (ii) (P^2+ Q^2+R^2)+P*P−QR=30.. also it is given P^2−QR=10.. so (P^2+ Q^2+R^2)+10=30.. or (P^2+ Q^2+R^2)=20.. C..

NOTE.. the moment we see .. three equations containing single term of square and we have to find sum of these three terms, In moost likelihood we will have to add the three terms... second point is when we see R≠Q, we can choose two equations where we get a term Q-R.. REST ofcourse is calculation
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