Question

(4+√15)^3÷2-(4-√15)^3÷2=√6.z
(x^3)1/2-(y^3)^1/2=√6.z
Squaring both the sides
x^3+y^3-2(xy)^3/2=6z^2
As we know the formula of a^3+b^3 i.e
(x+y)(x^2+y^2-xy) -2(xy)^3/2=6z^2
Now solve for LHS only
We know that x=(4+√15) where y=(4-√15)
Thus it implies:-
x+y= 8
(8)(x^2+y^2+2xy-3xy)- 2(xy)^3/2=6z^2
(8){(x+y)^2-3xy}-2(xy)^3/2=6z^2
(8){(8)^2-3xy}-2(xy)^3/2=6z^2
xy=(4+√15)(4-√15)=(4)^2-(√15)^2= 16-15=1
8(64-3)-2*1=6z^2
8*61-2=488-2=486=6z^2
z^2=486÷6
z=√81
z=- or +(9)

Answer 1

Answer:

z = ± 9

Step-by-step explanation

(4 + √15)^(3/2)  -  (4 + √15)^(3/2)) =  √6.z

Squaring both sides

(4 + √15)³ + (4 - √15)³ - 2(4 + √15)^(3/2) * (4 + √15)^(3/2)  = 6z²

using below

'(a + b)³ = a³ + b³ + 3a²b + 3ab²

(a-b)³ = a³ - b³ - 3a²b + 3ab²

(a + b)^(3/2) (a -b)^(3/2) = (a² - b²)^(3/2)

=> 64 + 15√15 + 48√15 + 180 + 64 - 15√15 - 48√15 + 180 - 2(16 - 15)^(3/2) = 6z²

=> 488 - 2(1)^(3/2) = 6z²

=> 488 - 2 = 6z²

=> 486 = 6z²

=> z² = 81

=> z = ± 9