find the value of a^3+b^3+c^3-3ab c when a+b+c=8 and ab+bc+ca= 25
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a + b + c = 8
ab + bc + ca = 25
_______ [GIVEN]
a³ + b³ + c³ - 3abc
_______ [FIND]
Solution:
=> a + b + c = 8
• Do squaring on both sides.
=> (a + b + c)² = (8)²
=> a² + b² + c² + 2ab + 2bc + 2ca = 64
=> a² + b² + c² + 2(ab + bc + ca) = 64
=> a² + b² + c² + 2(25) = 64
=> a² + b² + c² + 50 = 64
=> a² + b² + c² = 64 - 50
=> a² + b² + c² = 14
______________________________
We have to find a³ + b³ + c³ - 3abc
=> a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
=> a³ + b³ + c³ - 3abc = (a + b + c) [a² + b² + c² -(ab + bc + ca)]
=> a³ + b³ + c³ - 3abc = (8) [14 - 25]
=> a³ + b³ + c³ - 3abc = 8 (-11)
______________________________
a³ + b³ + c³ - 3abc = - 88
_____________ [
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