Q10. If [tex]a+b+c=5[/tex], [tex]ab+bc+ca=12[/tex] and [tex]abc=-1[/tex], find [tex]a^3+b^3+c^3[/tex].
Answer (1)
We are given a + b + c = 5 , ab + b c + c a = 12 , and ab c = − 1 .
Use the identity ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + b c + c a ) to find a 2 + b 2 + c 2 = 1 .
Use the identity a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) to find a 3 + b 3 + c 3 .
Substitute the values to get a 3 + b 3 + c 3 = − 58 , so the final answer is − 58 .
Explanation
Understanding the Problem We are given that a + b + c = 5 , ab + b c + c a = 12 , and ab c = − 1 . Our goal is to find the value of a 3 + b 3 + c 3 .
Using a Key Identity We will use the identity a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) to find a 3 + b 3 + c 3 .
Finding a 2 + b 2 + c 2 First, we need to find a 2 + b 2 + c 2 . We know that ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + b c + c a ) Substituting the given values, we have 5 2 = a 2 + b 2 + c 2 + 2 ( 12 ) 25 = a 2 + b 2 + c 2 + 24 a 2 + b 2 + c 2 = 25 − 24 = 1
Finding a 3 + b 3 + c 3 Now we can substitute the values of a + b + c , ab + b c + c a , ab c , and a 2 + b 2 + c 2 into the identity a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) a 3 + b 3 + c 3 − 3 ( − 1 ) = ( 5 ) ( 1 − 12 ) a 3 + b 3 + c 3 + 3 = ( 5 ) ( − 11 ) a 3 + b 3 + c 3 + 3 = − 55 a 3 + b 3 + c 3 = − 55 − 3 = − 58
Final Answer Therefore, a 3 + b 3 + c 3 = − 58 .
Examples
Understanding the relationship between the sums of powers of variables is useful in various fields, such as physics and engineering. For example, in mechanics, the kinetic energy of a system of particles can be expressed in terms of the sum of squares of velocities. Similarly, in electrical engineering, power dissipation in resistors can be related to the sum of squares of currents. These relationships help simplify complex calculations and provide insights into the behavior of systems.
