please answer the 29th question
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a + b + c = 5 (given)
ab + bc + ca = 10 (given)
so,
by using,a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)
(a + b + c)² = a² + b² + c² +2(ab + bc+ca)
(5)² -2×10 = a² + b² + c²
25-20=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 that a³+b³+c³-3abc= -25 .
ab + bc + ca = 10 (given)
so,
by using,a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)
(a + b + c)² = a² + b² + c² +2(ab + bc+ca)
(5)² -2×10 = a² + b² + c²
25-20=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 that a³+b³+c³-3abc= -25 .
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