Square of Difference: Using multiple rules

To solve this problem, we'll follow these steps:

• Step 1: Expand $(x+y)^2$
• Step 2: Expand $(x-y)^2$
• Step 3: Rearrange and simplify the entire expression

Now, let's work through each step:
Step 1: We expand $(x+y)^2$ using the formula for the square of a sum:
$(x+y)^2 = x^2 + 2xy + y^2$.

Step 2: We expand $(x-y)^2$ using the formula for the square of a difference:
$(x-y)^2 = x^2 - 2xy + y^2$.

Step 3: Substitute these expansions back into the original expression: $(x+y)^2-(x-y)^2+(x-y)(x+y)$ becomes: $(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) + (x-y)(x+y).$

First, simplify $(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)$:
– $x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xy.$

Next, consider $(x-y)(x+y)$:
By using the identity for difference of squares: $(x-y)(x+y) = x^2 - y^2$.

Thus, combining our results gives:
$4xy + x^2 - y^2 = x^2 + 4xy - y^2.$

Therefore, the solution to the problem is $x^2 + 4xy - y^2$.

To solve this problem, we'll follow these steps:

• Step 1: Expand $(x+3)^2$.
• Step 2: Expand $(x-3)^2$.
• Step 3: Simplify the expression by combining like terms.

Now, let's work through each step:
Step 1: Expand $(x+3)^2$ using the formula for the square of a sum:

$(x+3)^2 = x^2 + 2 \cdot x \cdot 3 + 3^2 = x^2 + 6x + 9$

Step 2: Expand $(x-3)^2$ using the formula for the square of a difference:

$(x-3)^2 = x^2 - 2 \cdot x \cdot 3 + 3^2 = x^2 - 6x + 9$

Step 3: Add the expanded expressions together and simplify:

$(x+3)^2 + (x-3)^2 = (x^2 + 6x + 9) + (x^2 - 6x + 9)$

$(x^2 + 6x + 9) + (x^2 - 6x + 9) = 2x^2 + 0x + 18 = 2x^2 + 18$

Therefore, the solution to the problem is $2x^2 + 18$.

To solve this problem, we will follow these steps:

• Step 1: Expand $(a+3b)^2$.
• Step 2: Expand $(3b-a)^2$.
• Step 3: Subtract the result of step 2 from step 1.

Now, let's work through the calculations:

Step 1: Expand $(a+3b)^2$
Using the formula $(x + y)^2 = x^2 + 2xy + y^2$, we let $x = a$ and $y = 3b$ to get:
$(a+3b)^2 = a^2 + 2 \cdot a \cdot 3b + (3b)^2$
$= a^2 + 6ab + 9b^2$.

Step 2: Expand $(3b-a)^2$
Again using the squaring formula, letting $x = 3b$ and $y = -a$, we have:
$(3b-a)^2 = (3b)^2 - 2 \cdot 3b \cdot a + a^2$
$= 9b^2 - 6ab + a^2$.

Step 3: Perform the subtraction
We subtract the expansion of $(3b-a)^2$ from $(a+3b)^2$:
$(a^2 + 6ab + 9b^2) - (9b^2 - 6ab + a^2)$
$= a^2 + 6ab + 9b^2 - 9b^2 + 6ab - a^2$
= $12ab$.

The solution to the problem is $12ab$, which corresponds to choice 2.

Let's examine the given equation:

$(x+3)^2=(x-3)^2$First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the resulting simplified equation:

$(x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.

To solve this problem, let's expand and simplify each side of the given equation:

• Step 1: Expand the left side:
  • Expand $(x + 3)^2 = x^2 + 6x + 9$
  • Expand $(2x - 3)^2 = 4x^2 - 12x + 9$
  Combine them to get $x^2 + 6x + 9 + 4x^2 - 12x + 9 = 5x^2 - 6x + 18$.
• Step 2: Expand the right side:
  • Expand $5x(x - \frac{3}{5}) = 5x^2 - 3x$
• Step 3: Set the equations from both sides equal and simplify: $5x^2 - 6x + 18 = 5x^2 - 3x$
• Step 4: Subtract $5x^2$ from both sides: $-6x + 18 = -3x$
• Step 5: Simplify to: $-6x + 3x = -18$ or equivalently $-3x = -18$
• Step 6: Solve for $x$: $x = \frac{-18}{-3} = 6$

Therefore, the solution to the problem is $x = 6$.

Let's solve this problem by simplifying each component separately:

First, simplify $(\frac{x}{3} - 4)^2$ using the square of a difference formula:

$(\frac{x}{3} - 4)^2 = \left(\frac{x}{3}\right)^2 - 2 \times \frac{x}{3} \times 4 + 4^2$ This becomes:

$= \frac{x^2}{9} - \frac{8x}{3} + 16$

Next, simplify $x(\frac{\sqrt{8x}}{3} + 2)(\frac{\sqrt{8x}}{3} - 2)$ using the difference of squares formula:

$(\frac{\sqrt{8x}}{3} + 2)(\frac{\sqrt{8x}}{3} - 2) = \left(\frac{\sqrt{8x}}{3}\right)^2 - 2^2$ Simplify further:

$= \frac{8x}{9} - 4$

Including the factor of $x$, we have:

$x \left(\frac{8x}{9} - 4\right) = \frac{8x^2}{9} - 4x$

Combine the results from both parts:

$\left(\frac{x^2}{9} - \frac{8x}{3} + 16\right) + \left(\frac{8x^2}{9} - 4x\right)$

Simplify by combining like terms:

$= \frac{x^2}{9} + \frac{8x^2}{9} - \frac{8x}{3} - 4x + 16 = x^2 - \left(\frac{8x}{3} + 4x\right) + 16 = x^2 - \left(\frac{8x + 12x}{3}\right) + 16 = x^2 - \frac{20x}{3} + 16$

Therefore, after simplifying, the expression becomes $\boldsymbol{x^2 - \frac{20x}{3} + 16}$.

The final solution is: $x^2 - \frac{20x}{3} + 16$.

To solve this problem, let's follow a detailed approach:

• First, expand both squares:
• $\left(\frac{2}{3} + \frac{m}{4}\right)^2 = \left(\frac{2}{3}\right)^2 + \frac{2 \times 2}{3 \times 4} \times m + \left(\frac{m}{4}\right)^2 = \frac{4}{9} + \frac{m}{6} + \frac{m^2}{16}$
• $\left(\frac{m}{4} - \frac{2}{3}\right)^2 = \left(\frac{m}{4}\right)^2 - \frac{2 \times m}{12} + \left(\frac{2}{3}\right)^2 = \frac{m^2}{16} - \frac{m}{6} + \frac{4}{9}$
• Now substitute and simplify into the expression:

• Expression becomes: $\left( \frac{4}{9} + \frac{m}{6} + \frac{m^2}{16} \right) - \frac{4}{3} - \left( \frac{m^2}{16} - \frac{m}{6} + \frac{4}{9} \right)$

• Observe that $\frac{m^2}{16} - \frac{m^2}{16}$ cancels out.
• Now simplify the remaining terms: $\frac{4}{9} + \frac{m}{6} - \frac{4}{3} + \frac{m}{6} - \frac{4}{9} = \frac{2m}{6} - \frac{4}{3}$
• Use common denominators to combine final terms:
• $\frac{2m}{6} = \frac{m}{3}$, $\frac{4}{3} = \frac{12}{9}$, resulting in: $\frac{m}{3} - \frac{12}{9}$
• Recognize that this is a difference of squares:
• $\frac{m}{3} - \frac{12}{9} = \frac{(\sqrt{2m} + 2)(\sqrt{2m} - 2)}{3}$

Therefore, the simplified expression is given by the choice: $\frac{(\sqrt{2m}+2)(\sqrt{2m}-2)}{3}$.