Question

if a + b + C =5 and AB + BC + CA =10 ,then prove that
${a}^{3} + {b}^{3} + {c}^{3} - 3abc = - 25$
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Answer 1

Step-by-step explanation:

a + b + c = 5

Squaring on both sides

(a+b+c)² = 5²

a² + b² + c² +2ab + 2bc + 2ac = 25

a² + b² + c² + 2(ab+bc+ac)=25

a² + b² + c² +2(10)=25 (AB + BC + CA =10)

a² + b² + c²=25-20

a² + b² + c²=5

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)

(5) [5-(ab+bc+ac)]

(5) [5-(10)]

5×-5=-25

hence proved