Question

If a, b and c are integers such that(cube root 4+cube root2-2)(cube root 4a+cube root 2b+c)=20 then one of the following is true? 1)a+b-c = 10 2) a-b+c=10 3) a+b=2c 4) a+b+c = 16​

Answer 1

Given : a , b and c are integers such that (³√4 + ³√2 - 2)(³√4a + ³√2b + c) = 20

To check : which one of the following is true ?

1. a + b - c = 10
2. a - b + c = 10
3. a + b = 2c
4. a + b + c = 16

solution : let (³√4 + ³√2 - 2) A = 20

⇒A = 20/(³√4 + ³√2 - 2)

= 20/(³√4 + ³√2 - 2) × (³√4 + ³√2 + 2)/(³√4 + ³√2 + 2)

= 20(³√4 + ³√2 + 2)/{(³√4 + ³√2)² - 2²}

= 20(³√4 + ³√2 + 2)/{(4⅓)² + (2⅓)² + 4 - 4}

= 20(³√4 + ³√2 + 2)/(4⅔ + 4⅓)

= (³√4 + ³√2 + 2) × 20/(4⅔ + 4⅓) ......(1)

we know, a³ + b³ = (a² + b² - ab)(a + b)

let a = 4⅔ , b = 4⅓

so, (4⅔)³ + (4⅓)³ = (4⅔ + 4⅓ - 4)(4⅔ + 4⅓)

⇒20/(4⅔ + 4⅓) = (4⅔ + 4⅓ - 4) , putting it in equation (1) we get,

A = (³√4 + ³√2 + 2)(4⅔ + 4⅓ - 4)

= 2³√2 - 2

= ³√4 × 0 + ³√(2 × 8) + (-2)

on comparing with (³√4a + ³√2b + c)

a = 0, b = 8 , c = -2

now , a + b - c = 0 + 8 + 2 = 10

a - b + c = 0 - 8 - 2 = -10

a + b = 0 + 8 ≠ 2 × 2 = 2c

a + b + c = 0 + 8 - 2 = - 6

Therefore option (1) is correct choice.