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Given that a+b=9 and a2+b2=71, find the value of (ab)3

Mathematics

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Answer

125\text{125}

Step-by-step solution

4 steps

  1. Use the algebraic identity for (a+b)2(a+b)^2.

    We know that (a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab.

  2. Substitute the given values into the identity.

    We are given a+b=9a+b=9 and a2+b2=71a^2+b^2=71.

    (9)2=71+2ab(9)^2 = 71 + 2ab 81=71+2ab81 = 71 + 2ab

  3. Solve for abab.

    Subtract 71 from both sides:

    81−71=2ab81 - 71 = 2ab 10=2ab10 = 2ab

    Divide by 2:

    ab=102ab = \frac{10}{2} ab=5ab = 5

  4. Calculate the value of (ab)3(ab)^3.

    Substitute the value of abab into (ab)3(ab)^3:

    (ab)3=(5)3(ab)^3 = (5)^3 (ab)3=5×5×5(ab)^3 = 5 \times 5 \times 5 (ab)3=125(ab)^3 = 125

    The value of (ab)3(ab)^3 is 125\boxed{\text{125}}.

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