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anonymous

  • one year ago

The sum of the squares of three negative numbers p,q,r equals the sum of the products of all possible pairs of the three numbers. Which of the following equals p3qr2 ? 1) (pq + qr + rp)3 2) q2(p3r + rp3)/2

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    we're given : \(p^2+q^2+r^2 = pq=qr=rp\) so, \(p^3qr^2 = pq*(pr)^2 = (p^2+q^2+r^2)^3\)

  3. anonymous
    • one year ago
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    Right answer id option 2 @ganeshie8

  4. TuringTest
    • one year ago
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    "The sum of the squares of three negative numbers p,q,r equals the *sum* of the products of all possible pairs of the three numbers." I think we are given that\[p^2+q^2+r^2=pq+qr+rp\]I could be wrong; I'm a bit rusty...

  5. hartnn
    • one year ago
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    yes, ^^

  6. anonymous
    • one year ago
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    Which of the following equals p3qr2 ?@TuringTest

  7. TuringTest
    • one year ago
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    I'm pretty bad at these problems, but\[p^2=pq+qr+rp-q^2-r^2\]\[r^2=pq+qr+rp-q^2-p^2\]\[pq=p^2+q^2+r^2-qr-rp\]multiplying all that out should give an answer, but I'm sure there's a smarter way to do it.

  8. anonymous
    • one year ago
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    Thank you :)

  9. TuringTest
    • one year ago
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    You're welcome

  10. hartnn
    • one year ago
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    something is off.. the answer is q2(p3r + rp3)/2 <<p^3 r and r p^3 are same!!>> which means p^3 q r^2 = q^2(p^3 r + rp^3)/2 holds true p^3 and q can be factored out from the right and gets cancelled, r^2 = q (r+r)/2 = rq r= q holds true ...

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