Question

simplify =( x+y3y)whole cube + (3x+y) whole square​

Answer 1

Answer:

$(x+3y^{2})^{3}+(3x+y)^{2}$ =  $x^{3} +27y^{6} +9xy^{2} (x+3y^{2})+9x^{2} +6xy+y^{2}$

Step-by-step explanation:

Given : The expression is $(x+3y^{2})^{3}+(3x+y)^{2}$

To find : The value of given expression.

Solution :

• The given expression is, $(x+3y^{2})^{3}+(3x+y)^{2}$
• We have to simplify the given expression and find out value of the expression.
• To simplify the expression,we should know two identities,

      $(a+b)^{3} =a^{3} +b^{3} +3ab(a+b)$ and

      $(a+b)^{2} = a^{2} +2ab+b^{2}$

• Now, applying these two identities in the given equation, we get

$(x+3y^{2})^{3}+(3x+y)^{2}$ = $x^{3} +(3y^{2}) ^{3} +3x*3y^{2} (x+3y^{2})+(3x)^{2} +2*3x*y+y^{2}$

                             = $x^{3} +27y^{6} +9xy^{2} (x+3y^{2})+9x^{2} +6xy+y^{2}$

• ∴ $(x+3y^{2})^{3}+(3x+y)^{2}$ =  $x^{3} +27y^{6} +9xy^{2} (x+3y^{2})+9x^{2} +6xy+y^{2}$