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To solve the algebraic expression $({{x}^{2}}-{{y}^{2}})\times (x+2y)$.

Last updated date: 19th Jul 2024
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Hint: While solving this problem it is important to know the multiplication of algebraic expressions. That is, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$ (where, x is a algebraic variable, a and b are any integers).

Complete step-by-step solution:
To solve any algebraic expression involving basic arithmetic operations like addition, subtraction, multiplication and division, we follow the basic procedures as below-
For addition and subtraction, ax + bx + cy – dy = (a+b) x + (c-d) y
We can combine constants (in this case a and b for x and c and d for y) for the same variable (as shown above).
In case of multiplication and division,
(ax+by)$\times $(cx+dy)
First multiplying ax with (cx+dy) and then by with (cx+dy), we get,
= [(ax)$\times $(cx+dy)] + [(by)$\times $(cx+dy)]
=[ac${{x}^{2}}$+adxy] + [bcxy + bd${{y}^{2}}$]
Now, simply following rules of addition, we get,
= ac${{x}^{2}}$+ (ad+bc)xy + bd${{y}^{2}}$
Now, applying these rules to solve the question given,
=$({{x}^{2}}-{{y}^{2}})\times (x+2y)$
Now multiplying ${{x}^{2}}$ with (x+2y) and then $(-{{y}^{2}})$ with (x+2y), we get,
= [${{x}^{2}}$$\times $ (x+2y)] + [$(-{{y}^{2}})$$\times $(x+2y)]
Now, using the property, ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$, we get,
=[${{x}^{2+1}}+2{{x}^{2}}y$] + [$-{{y}^{2}}x-2{{y}^{2+1}}$]
Hence, the solution to the algebraic expression of the question is ${{x}^{3}}+2{{x}^{2}}y-{{y}^{2}}x-2{{y}^{3}}$.
Note: Alternative way to solve the question is by multiplying (x+2y) by $({{x}^{2}}-{{y}^{2}})$. That is,
(x+2y)$\times $ $({{x}^{2}}-{{y}^{2}})$
That is this time we multiply x by $({{x}^{2}}-{{y}^{2}})$ and then add that by the expression we get by multiplying (2y) by $({{x}^{2}}-{{y}^{2}})$, thus,
[(x)$\times $ $({{x}^{2}}-{{y}^{2}})$]+[2y$\times $$({{x}^{2}}-{{y}^{2}})$]
  & =[{{x}^{3}}-x{{y}^{2}}]+[2y{{x}^{2}}-2{{y}^{3}}] \\
 & ={{x}^{3}}-x{{y}^{2}}+2y{{x}^{2}}-2{{y}^{3}} \\
Which is the same expression we got in the solution.