Question

If a + b + c = 5, a2 + b2 + c2 = 27, and a3 + b3 + c3 = 125, then the value of 4abc is:

Answer 1

Step-by-step explanation:

Given If a + b + c = 5, a2 + b2 + c2 = 27, and a3 + b3 + c3 = 125, then the value of 4abc is

• We know that a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)
• (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
•  5^2 = 27 + 2(ab + bc + ca)
• 25 – 27 = 2(ab + bc + ca)
• -2 / 2 = ab + bc + ca
• So ab + bc + ca = - 1-----------1
• Now a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)
•                125 – 3abc = 5(27 – (- 1)
•                 125 – 3abc = 5(28)
•                    125 – 3abc = 140
•                   125 – 140 = 3abc
•          So 3abc = - 15
•     Or abc = - 5
• Multiplying by 4 we get

   4abc = - 20

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Answer 2

4abc = -20

Step-by-step explanation:

Given data

a+b+c = 5

a²+ b² + c² = 27

a³ + b³ + C³ = 125

To find - 4 abc

(a+b+c)² = a² + b² + c² + 2ab + 2bc +2ca  ------------> 1

Substitute the values of (a+b+c) and a²+b²+c² in the equation 1

(5)² = 27 + 2(ab + bc+ ca)

25 = 27 +  2(ab + bc+ ca)

-2 =  2(ab + bc+ ca)

Eliminate 2 on both sides of the equation

(ab + bc+ ca) = -1   -------------->2

(a³ + b³ +c³) = (a+b+c)(a² + b² +c² -ab -bc-ca) ------------>3

Substitute a+b+c = 5, a²+ b² + c² = 27, a³ + b³ + C³ = 125 in the above equation

(125- 3abc) = 5 (27 - ( -1))

(125- 3abc) = 135 + 5

(125- 3abc) = 140

-3abc = 15

abc = - 5

Multiply 4 on the both sides of the above equation

4 abc = 4× (-5)

4 abc = -20

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1. https://brainly.in/question/4855440

2. https://brainly.in/question/3864104