Question

Please solve number 26

Answer 1

=>
(x^3+12x+6x^2+8)/(x^3+12x-6x^2-8) = (y^3+27y+9y^2+27)/(y^3+27y-9y^2-27)[by using componendo and dividendo)

=>
The numerators are in the form (a^3+b^3+3a^2b+3ab^2) and denominators are in the form (a^3-b^3-3a^2b+3ab^2).
we also know that [a^3+b^3+3a^2b+3ab^2 = (a+b)^3] and [a^3-b^3-3a^2b+3ab^2 = (a-b)^3]

=>
Now we can rewrite the equation as
(x+2)^3/(x-2)^3 = (y+3)^3/(y-3)^3

=>
take cube root on both sides we get
(x+2)/(x-2) = (y+3)/(y-3)

=>
Now apply componendo and dividendo rule
(x+2+x-2)/(x+2-x+2) = (y+3+y-3)/(y+3-y+3)

=>
2x/4 = 2y/6
x/y = 2/3

=>
therefore x:y = 2:3

don't hesitate to ask if you're having any doubts

Answer 2

Answer:

Step-by-step explanation:

By using componendo and dividendo:-

x³+12x+6x²+8=y³+27y+9y²+27

__________ ____________

x³+12x-6x²-8 y³+27y-9y²-27

Using LHS

(x)³+3x*2²+3*2(x)²+(2)³

___________________

(x)³+3x*(2)²-3*2(x)²+(x)²

=>(x+y)³

___ --------1

(x-y)³

Using RHS

(y)³+3*(3)²*y+3*(y)²*3+(3)³

___________________

(y)³+3*(3)²*y-3*(y)²*3-(3)³

=>(y+3)³

____ --------2

(y-3)³

By using 1 and 2

(x+2)³=(y+3)³

___ ___

(x-2)³ (y-3)³

=> (x+2)=(y+3)

___ ___

(x-2) (y-3)

By using componendo and dividendo:-

x = y

_ _

2 3

x = 2

_ _

y 3