20. Find the value of a3 + b3 + c3 - 3abc , if
Answers
Step-by-step explanation:
(i) Have Given :-
a + b + c = 14
a² + b² + c² = 68
To find :-
a³ + b³ + c³ – 3abc = ?
Solution :-
a + b + c = 14
On squaring both sides :-
(a + b + c)² = (14)²
a² + b² + c² + 2ab + 2bc + 2ca = 196
( a² + b² + c² ) + (2ab + 2bc + 2ca) = 196
68 + 2( ab + bc + ca) = 196 (∵ a² + b² + c² = 68)
2( ab + bc + ca) = 196 – 68
2( ab + bc + ca) = 128
ab + bc + ca = 128/2
ab + bc + ca = 64
Formula :-
(a³+b³+c³) – 3abc = (a+b+c) {a²+b²+c²–ab–bc–ca}
a³ + b³ + c³– 3abc = 14{68 – (ab + bc + ca)}
a³ + b³ + c³– 3abc = 14{ 68 – 64}
a³ + b³ + c³– 3abc = 14{ 4 }
a³ + b³ + c³– 3abc = 56
(ii) Have Given :-
a + b + c = 9
ab + bc + ca = 26
To find :-
a³ + b³ + c³ – 3abc = ?
Solution :-
a + b + c = 9
On squaring both sides :-
(a + b + c)² = (9)²
a² + b² + c² + 2ab + 2bc + 2ca = 81
(a² + b² + c²) + 2( ab + bc + ca) = 81
a² + b² + c² + 2(26) = 81
a² + b² + c² + 52 = 81
a² + b² + c² = 81 – 52
a² + b² + c² = 29
Formula :-
(a³+b³+c³) – 3abc = (a+b+c) {a²+b²+c²–ab–bc–ca}
a³ + b³ + c³– 3abc = 9 {29 – (ab + bc + ca)}
a³ + b³ + c³– 3abc = 9 {29 – 26}
a³ + b³ + c³– 3abc = 9 {3}
a³ + b³ + c³– 3abc = 27
I hope it is helpful
Answer:
thankyou so much nilpriya for thanking my answers..