Question

If 9y² + 1/y² =3 then find the value of 27y³ + 1/y³

Answer 1

Step-by-step explanation:

hope it will help thankful

Answer 2

The value of $27y^3+\dfrac{1}{y^3}$ is equal to 0.

Step-by-step explanation:

Given that,

$9y^2+\dfrac{1}{y^2}=3$........(1)

Using $(a+b)^2$ in $(3y+\dfrac{1}{4})^2$, so,

$(3y+\dfrac{1}{y})^2=9y^2+\dfrac{1}{y^2}+2\times 3y\times \dfrac{1}{y}\\\\(3y+\dfrac{1}{y})^2=9y^2+\dfrac{1}{y^2}+6$

Using equation (1) in above equation as :

$(3y+\dfrac{1}{y})^2=9y^2+\dfrac{1}{y^2}+6\\\\(3y+\dfrac{1}{y})^2=3+6\\\\3y+\dfrac{1}{y}=3..................(2)$  

The formula of $(a+b)^3$ is :

$(a+b)^3=a^3+b^3+3ab(a+b)\\\\(3y+\dfrac{1}{y})^3=(3y)^3+(\dfrac{1}{y})^3+3\times 3y\times \dfrac{1}{y}(3y+\dfrac{1}{y})$

Using equation (2) in above equation :

$(3y+\dfrac{1}{y})^3=(3y)^3+(\dfrac{1}{y})^3+3\times 3y\times \dfrac{1}{y}(3y+\dfrac{1}{y})\\\\3^3=(3y)^3+(\dfrac{1}{y})^3+27\\\\27y^3+\dfrac{1}{y^3}=0$

So, the value of $27y^3+\dfrac{1}{y^3}$ is equal to 0.

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Algebra