If a+b =11 and ab=28, find the value of a^3+b^3
krish02:
another process is we can also substitute a=7 and b=4 because product of ab=28 and their sum a+b=11
so that a^3+b^3=7^3+4^3=407
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Answered by
3
ab = 28 --------(1)
a+b = 11 --------(2)
On squaring both sides, we get
a^2 + b^2 +2ab = 121
=> a^2 +b^2 + 56 = 121
=> a^2 +b^2 = 65 -------(3)
Now,
a^3 +b^3 = (a+b) (a^2 +b^2 - ab)
= 11 (65 - 28)
= 11 * 37
= 407
a+b = 11 --------(2)
On squaring both sides, we get
a^2 + b^2 +2ab = 121
=> a^2 +b^2 + 56 = 121
=> a^2 +b^2 = 65 -------(3)
Now,
a^3 +b^3 = (a+b) (a^2 +b^2 - ab)
= 11 (65 - 28)
= 11 * 37
= 407
Thank you
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Answered by
2
a=7 and a=4 so that their sum is 11 and their product is 28 so that a^3+b^3=7^3+4^3=407
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