Simplifying the Expression: (x+y)^3-(x-y)^3-6y(x^2-y^2)=ky^3
This problem involves simplifying a given algebraic expression and finding the value of the constant 'k'. Let's break down the steps to reach the solution.
Expanding the Expression
Firstly, we need to expand the cubes of the binomial expressions using the formula: (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
Applying this to our expression: (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
Substituting these expansions into our original expression: (x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) - 6y(x^2 - y^2) = ky^3
Simplifying Further
Now, we simplify the expression by combining like terms: 6x^2y + 2y^3 - 6yx^2 + 6y^3 = ky^3
Combining the y^3 terms and cancelling out the x^2y terms, we get: 8y^3 = ky^3
Determining the Value of 'k'
Finally, to find the value of 'k', we simply divide both sides of the equation by y^3 (assuming y is not equal to zero): 8 = k
Therefore, the value of k is 8.
Conclusion
By expanding the binomial cubes and simplifying the expression, we have found that the value of 'k' in the equation (x+y)^3-(x-y)^3-6y(x^2-y^2)=ky^3 is 8.