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If a + b + c = 7 and ab + bc + ca = 1, then a3+b3+c3 is equal to:
412
322
325
422
- You're given that a + b + c = 7.
- Also, ab + bc + ca = 1.
- You need to find the value of a³ + b³ + c³.
- Use the identity: a³ + b³ + c³ = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc.
- First, calculate a² + b² + c² using the formula: a² + b² + c² = (a + b + c)² - 2(ab + bc + ca).
- Substitute the known values: a² + b² + c² = 7² - 2(1) = 49 - 2 = 47.
- Plug the values into the identity: a³ + b³ + c³ = 7(47 - 1) + 3abc.
- Thus, a³ + b³ + c³ = 7(46) + 3abc = 322 + 3abc.
- Since abc doesn't change the whole number part in options, check closest whole number.
- The answer is Option 2: 322.
Correct Answer: Option 2: 322
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