Question

Show that the following identities hold.

(a)$ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$

(b)width="208">a2+b22−a2−b22=4a2b2

Short answer

(a) So, the identity holds $ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$

(b) So, the identity holds${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$

Step-by-step solution

Part a. Step 1. Apply the special product formula.

If A, B and C are any real numbers or algebraic expressions, then

${\left(A+B\right)}^2=A^2+2AB+B^2$

width="174">A−B2=A2−2AB+B2

Part a. Step 2. Analyze the given identity.

Given$ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$

Here Left Hand Side (LHS) is$ab$

Right Hand Side (RHS) is$\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$

Part a. Step 3. Simplify Right Hand Side.

$\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$is the RHS

Simplify the RHS to get:

$\begin{array}{l}\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=\frac{1}{2}\left[a^2+2ab+b^2-a^2-b^2\right]\\ \Rightarrow \frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=\frac{1}{2}\left[a^2-a^2+2ab+b^2-b^2\right]\\ \Rightarrow \frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=\frac{1}{2}\left[2ab\right]\\ \Rightarrow \frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=ab\end{array}$

Part a. Step 4. Conclusion.

Since RHS is simplified as $\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]=ab$, which is equal to LHS.

Therefore,$ab=\frac{1}{2}\left[{\left(a+b\right)}^2-\left(a^2+b^2\right)\right]$

Hence proved

Part b. Step 1. Apply the special product formula.

If A, B and C are any real numbers or algebraic expressions, then

${\left(A+B\right)}^2=A^2+2AB+B^2$

${\left(A-B\right)}^2=A^2-2AB+B^2$

Part b. Step 2. Analyze the given identity.

Given${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$

Here Left Hand Side (LHS) is${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2$

Right Hand Side (RHS) is$4a^2{b}^2$

Part b. Step 3. Simplify Left Hand Side.

${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2$is the LHS

Simplify the LHS to get:

$\begin{array}{l}{\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2={\left(a^2\right)}^2+2a^2{b}^2+{\left(b^2\right)}^2-\left[{\left(a^2\right)}^2-2a^2{b}^2+{\left(b^2\right)}^2\right]\\ \Rightarrow {\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2={\left(a^2\right)}^2+2a^2{b}^2+{\left(b^2\right)}^2-{\left(a^2\right)}^2+2a^2{b}^2-{\left(b^2\right)}^2\\ \Rightarrow {\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2={\left(a^2\right)}^2-{\left(a^2\right)}^2+2a^2{b}^2+2a^2{b}^2+{\left(b^2\right)}^2-{\left(b^2\right)}^2\\ \Rightarrow {\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2\end{array}$

Part b. Step 4. Conclusion.

Since LHS is simplified as ${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$, which is equal to RHS.

Therefore,${\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2=4a^2{b}^2$

Hence proved