Question

if x= 2-root 3,y=root 3 -root 2 and z =root 7 -root 4 find the value of x^3+y^3+z^3

Answer 1

Answer:

x³ + y³ + z³ = 11(√3-√2-√7) - 50

Step-by-step explanation:

x = 2√3

y = √3-√2

z = √7-√4 = √7 - 2

x³ + y³ + z³

= (2√3)³ + (√3-√2)³ + (√7-2)³

= 14√3 + (3√3 - 2√2 - 3√6(√3-√2)) + (7√7 - 8 - 6√7(√7-2))

= 14√3 + 3√3 - 2√2 - 3√18 - 3√12 + 7√7 - 8 - 42 - 12√7

= 14√3 + 3√3 - 2√2 - 9√2 - 6√3 + 7√7 - 8 - 42 - 12√7

= 11√3 - 11√2 - 11√7 - 50

= 11 × (√3-√2-√7) - 50

Answer 2

The correct question is,

The correct question is, If x = 2-√3, y = √3-√7 and z = √7-√4 find the value of x³ + y³ + z³

Answer:

3 (3√3 + √7 - 8)

Explanation:

x = 2-√3

y = √3-√7

z = √7-√4

If x + y + z = 0,

x³ + y³ + z³ = 3xyz

Here,

2-√3 + √3-√7 + √7-√4 = 0

So,

x³ + y³ + z³ = 3xyz

= 3 (2-√3) (√3-√7) (√7-√4)

= 3 (2√3 - 2√7 - 3 + √21) (√7 -2)

=3(2√21-4√3-14-4√7-3√7+6+7√3-2√21)

= 3 (3√3 + √7 - 8)

Hence, the answer is 3 (3√3 + √7 - 8).

Hope it helps!...