If a+b+ c =9, then ab+bc+ca=23, then a³+b³ + c³ – 3abc =
A. 108
B. 207
C. 669
D. 729
Answers
Given : a + b + c = 9 and ab + bc + ca = 23
By using an algebraic identity :
a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
a³ + b³ + c³ – 3abc = (a + b + c) {(a² + b² + c²) - (ab + bc + ca)} ……..(1)
By using an algebraic identity :
(a + b + c)² = a² + b² + c² + 2ab + 2 bc + 2 ca)]
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(9)² = a² + b² + c² + 2(23)
81 = a² + b² + c² + 46
a² + b² + c² = 81 - 46
a² + b² + c² = 35
Now putting a + b + c = 9 , ab + bc + ca = 23 and a² + b² + c² = 35 in eq 1,
a³ + b³ + c³ – 3abc = (9) {(35) - (23)}
a³ + b³ + c³ – 3abc = 9 {35 - 23}
a³ + b³ + c³ – 3abc = 9 × 12
a³ + b³ + c³ – 3abc = 108
Hence, the value of a³ + b³ + c³ – 3abc is 108.
Option (A) 108 is correct.
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If a+b+ c =9 and ab + bc + ca = 23, then a²+b² + c²=
A. 35
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Answer:
Step-by-step explanation:
a + b + c)² = a² + b² + c² + 2ab + 2 bc + 2 ca)]
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(9)² = a² + b² + c² + 2(23)
81 = a² + b² + c² + 46
a² + b² + c² = 81 - 46
a² + b² + c² = 35
Now putting a + b + c = 9 , ab + bc + ca = 23 and a² + b² + c² = 35 in eq 1,
a³ + b³ + c³ – 3abc = (9) {(35) - (23)}
a³ + b³ + c³ – 3abc = 9 {35 - 23}
a³ + b³ + c³ – 3abc = 9 × 12
a³ + b³ + c³ – 3abc = 108