Question

20. Find the value of a3 + b3 + c3 - 3abc , if ​

Answer 1

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 2

Answer:

thankyou so much nilpriya for thanking my answers..