Learn and know the **formula of a minus b whole square** in algebra. a minus b whole square is also called as an identity. This formula is one of the most important formula in algebra.

Along with the formula mentioning I have given **derivation of (a-b) ^{ 2} **formula also. We can derive the formula in two different methods. First method is from the regular process that is give below. The second method is from (a+b)

^{ 2}formula we can

**derive (a-b)**For that what we need to do is in + sign you replace it with – sign.

^{ 2}formula.**Derivation of a minus b whole square formula:**

(a-b)^{ 2} = (a-b) × (a-b)

= a× (a-b) – b× (a-b)

= a^{2} – ab –ab + b^{2}

= a^{2} -2ab + b^{2}

Therefore, (a-b)^{ 2} = a^{2} -2ab + b^{2}

**a minus b whole square formula as follows:**

**a minus b whole square** is equal to a square (a^{2}) minus (-) product of 2, a and b plus (+) b square (b^{2})

i.e. (a-b)^{ 2} = a^{2} -2ab + b^{2}

(a-b)^{ 2} + 2ab = a^{2} + b^{2}

Note:

**a minus b whole square** in terms of a plus b whole square formula.

(a-b)^{ 2} = (a+b)^{ 2} – 4ab

Example:

♦ **Find (a-b) ^{ 2} value if a = 7 and b = 3 by using the formula.**

Solution:

We know that,

(a-b)^{ 2} = a^{2} -2ab + b^{2}

(a-b)^{ 2} = 7^{2} -2.7.3 + 3^{2}

(a-b)^{ 2} = 49 – 42 + 9

(a-b)^{ 2} = 58 – 42

(a-b)^{ 2} = 16

♦ **Find a ^{2} + b^{2} value if a-b = 3 and ab = 5.**

Solution:

We know that

(a-b)^{ 2} + 2ab = a^{2} + b^{2}

(3)^{2} + 2. 5 = a^{2} + b^{2}

9 + 10 = a^{2} + b^{2}

19 = a^{2} + b^{2}

♦ **Find (a-b) ^{ 2} value if a+b = 6 and ab = 2.**

Solution:

We know that,

(a-b)^{ 2} = (a+b)^{ 2} – 4ab

(a-b)^{ 2} = (6)^{ 2} – 4.2

(a-b)^{ 2} = 36 – 8

(a-b)^{ 2} = 28

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