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Chapter 5: Problem 65

Find each product. $$(x+y)\left(x^{2}+3 x y+y^{2}\right)$$

Short Answer

Expert verified
So the product is \(x^{3} +4x^{2}y +4xy^{2} + y^{3}\)

Step by step solution

01

Distribute the first term of the binomial

Use the distributive property first with the first term, \(x\), of the binomial, \((x+y)\).This gives: \(x(x^{2}) +x(3xy) + x(y^{2}) = x^{3} + 3x^{2}y + xy^{2}\)
02

Distribute the second term of the binomial

Now, apply the distributive property again with the second term, \(y\), of the binomial. The second round of distribution gives: \(y(x^{2}) +y(3xy) + y(y^{2}) = x^{2}y + 3xy^{2} + y^{3}\)
03

Add the two expressions

Finally, combine the two expressions obtained by distribution:This gives: \(x^{3} +3x^{2}y +xy^{2} +x^{2}y + 3xy^{2} + y^{3} = x^{3} +4x^{2}y +4xy^{2} + y^{3}\)

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Most popular questions from this chapter

Use a vertical format to find each product. $$\begin{array}{r}2 x^{3}+x^{2}+2 x+3 \\\\\quad x+4 \\\\\hline\end{array}$$

Use a vertical format to find each product. $$\begin{aligned}&3 y^{3}+2 y^{2}+y+4\\\&y+3\end{aligned}$$

Perform the indicated operations. $$(y+1)\left(y^{2}-y+1\right)-(y-1)\left(y^{2}+y+1\right)$$

Simplify: \(24+8 \cdot 3+28 \div(-7)\).

The mad Dr. Frankenstein has gathered enough bits and pieces (so to speak) for \(2^{-1}+2^{-2}\) of his creature-to-be. Write a fraction that represents the amount of his creature that must still be obtained.
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