Question

if ab+bc+ca=10 and a²+b²+c²=44 find a³+b³+c³-3abc​..​

Answer 1

Answer:

We know ,

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

now ,

a + b + c = 5

ab + bc + ca = 10

(a + b + c)² = a² + b² + c² +2(ab + bc+ca)

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5

hence ,

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

=( 5)( 5 - 10) = 5 × (-5) = -25

hence proved

Step-by-step explanation:

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