Question

If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 3abc = 25.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

✨✨✨✨ANSWER...

FORMULA USED

$(a {}^{2} + b {}^{2} + c {}^{2} ) = (a + b + c) {}^{2} - 2(ab + bc + ca) \\ (putting \: the \: values \: \: we \: get...) \\ (a {}^{2} + b {}^{2} + c {}^{2} ) =(5) {}^{2} - 2(10) \\ (a {}^{2} + b {}^{2} + c {}^{2} ) =25 - 20 \\ (a {}^{2} + b {}^{2} + c {}^{2} ) =5$

$now \: .... \\ a {}^{3} + b {}^{3} + c {}^{} - 3abc = (a + b + c)((a {}^{2} + b {}^{2} + c {}^{2} - (ab + bc + ca)) \\ \\ putting \: the \: values \\ \\ a {}^{3} + b {}^{3} + c {}^{} - 3abc = (5)(5 - 10) \\ a {}^{3} + b {}^{3} + c {}^{} - 3abc = - 25$

• so it has been proved above.............hope this helps you