Question 201145: The sum of two numbers A and B is 3. The sum of their cubes is 36, What is the sum of their squares decreased by the product of the two numbers?
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! (1) A + B = 3,,,,,,,,B=(3-A)
(2) A^3 + B^3 =36
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A^3 + (3-A)^3 =36
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side problem (3-A)^3 = (3-A) (3-A)(3-A)
(3-A) ( 9-6A +A^2)
27 -27A +9A^2 -A^3
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subst in
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A^3 +(27 -27A +9A^2 -A^3) =36
9A^2 -27A -9 =0
9(A^2 -3A -1) =0
(A^2 -3A -1) =0
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To factor use quadratic eqn
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a=1,,,,,b=-3,,,,,c=-1
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A =[ -(-3) +/- sqrt((-3)^2 - 4 (1)(-1)] / 2(1)
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A= [3 +/- sqrt{ 9+4} ]/2
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A= [3 +/- 3.6]/2
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A= 3.3, -.3,,,,,
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B= 3-3.3 =-.3,,,,or B= 3-(-.3) = 3.3
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( A^2 +B^2) - AB =?
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{ (3.3)^2 +(-.3)^2 } - (3.3)(-.3) = ?
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11+10 = ?
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21=?
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