If (a + b - c) = 20, and a2+ b+ c2 = 152, find the value of a3 + b3 - c3 + 3abc.

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SSC CGL 2023 Tier-I Official Paper (Held On: 25 Jul 2023 Shift 4)
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  1. 480
  2. 720
  3. 640
  4. 560

Answer (Detailed Solution Below)

Option 4 : 560
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Detailed Solution

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Given:

(a + b - c) = 20;

a2+ b+ c2 = 152

Formula used:

a3 + b3 = (a + b)3 - 3ab × (a + b)

Calculation:

We have 3 variables in this equation and 2 equations are given.

then let c = 0

⇒ (a + b) = 20

⇒ a2 + b2 = 152

Now,

(a + b) = 20

Squaring both sides

⇒ a2 + b2 + 2ab = 400

⇒ 152 + 2ab = 400

⇒ 2ab = 400 - 152 = 248

⇒ ab = 248/2 = 124

a3 + b3 = (a + b)3 - 3ab × (a + b)

⇒ (20)3 - 3 × 124 × 20

⇒ 8000 - 7440 = 560

∴ The correct answer is 560.

Alternate MethodConcept:

Use algebraic identities to find the required expression.

Calculation

Use the identity for the sum of cubes:

We know, 

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

if c ⇒ - c, then

a3 + b3 - c3 + 3abc = (a + b - c){(a2 + b2 + c2 - ab - b(-c) - (-c)a)}

⇒ 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)} 

Given:

a + b - c = 20

a2 + b2 + c2 = 152 

Find ( ab + bc - ca):

⇒ (a + b - c)2 = a2 + b2 + c+ 2ab - 2bc - 2ca

⇒ 202 = 152 + 2ab - 2bc - 2ca

⇒ 400 = 152 + 2(ab - bc - ca) 

⇒ 248 = 2(ab - bc - ca) 

⇒ ab - bc - ca = 124 

Substitute in the expression:

a3 + b3 - c3 + 3abc

⇒(a + b - c){(a2 + b2 + c2 - (ab - bc - ca)} 

= 20 ×  (152 - 124)

= 20 × 28 

= 560 

∴ The value of a3 + b3 - c3 + 3abc is 560.

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