In this section, we are going to see the formula or expansion for (a + b + c)^{2}.

That is,

(a + b + c)^{2} = (a + b + c)(a + b + c)

(a + b + c)^{2} = a^{2} + ab + ac + ab + b^{2 }+ bc + ac + bc + c^{2}

**(a + b + c) ^{2} = a^{2} + b^{2 }+ c^{2 }+ 2ab + 2bc + 2ac**

**Example 1 : **

Expand :

(5x + 3y + 2z)^{2}

**Solution : **

(5x + 3y + 2z)^{2 }is in the form of (a + b + c)^{2}

Comparing (a + b + c)^{2} and (5x + 3y + 2z)^{2}, we get

a = 5x

b = 3y

c = 2z

Write the formula / expansion for (a + b + c)^{2}.

(a + b + c)^{2} = a^{2} + b^{2 }+ c^{2 }+ 2ab + 2bc + 2ac

Substitute 5x for a, 3y for b and 2z for c.

(5x + 3y + 2z)^{2} :

= (5x)^{2 }+ (3y)^{2 }+ (2z)^{2 }+ 2(5x)(3y) + 2(3y)(2z) + 2(5x)(2z)

(5x + 3y + 2z)^{2} = 25x^{2 }+ 9y^{2 }+ 4z^{2 }+ 30xy + 12yz + 20xz

So, the expansion of (5x + 3y + 2z)^{2 }is

25x^{2 }+ 9y^{2 }+ 4z^{2 }+ 30xy + 12yz + 20xz

**Example 2 : **

If a + b + c = 15 , ab + bc + ac = 25, then find the value of

a^{2} + b^{2} + c^{2}

**Solution : **

To get the value of (a^{2} + b^{2} + c^{2}), we can use the formula or expansion of (a + b + c)^{2}.

Write the formula / expansion for (a + b + c)^{2}.

(a + b + c)^{2} = a^{2} + b^{2 }+ c^{2 }+ 2ab + 2bc + 2ac

(a + b + c)^{2} = a^{2} + b^{2 }+ c^{2 }+ 2(ab + bc + ac)

Substitute 15 for (a + b + c) and 25 for (ab + bc + ac).

(15)^{2} = a^{2} + b^{2 }+ c^{2 }+ 2(25)

225 = a^{2} + b^{2 }+ c^{2 }+ 50

Subtract 50 from each side.

175 = a^{2} + b^{2 }+ c^{2}

So, the value of a^{2} + b^{2 }+ c^{2 }is 175.

To get formula / expansion for (a + b - c)^{2}, let us consider the formula / expansion for (a + b + c)^{2}.

The formula or expansion for (a + b + c)^{2 }is

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac**

In (a + b + c)^{2}, if c is negative, then we have

(a + b - c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find "c".

They are c^{2}, bc, ca.

Even if we take negative sign for "c" in c^{2}, the sign of c^{2} will be positive. Because it has even power 2.

The terms bc, ac will be negative. Because both "b" and "a" are multiplied by "c" that is negative.

Finally, we have

**(a + b - c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab - 2bc - 2ac**

**Example :**

Expand :

(x + 2y - z)^{2}

**Solution : **

(x + 2y - z)^{2 }is in the form of (a + b - c)^{2}

Comparing (a + b - c)^{2} and (x + 2y - z)^{2}, we get

a = x

b = 2y

c = z

Write the formula / expansion for (a + b - c)^{2}.

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

Substitute x for a, 2y for b and z for c.

(x + 2y - z)^{2 }:

= x^{2 }+ (2y)^{2 }+ z^{2 }+ 2(x)(2y) - 2(2y)(z) - 2(x)(z)

(x + 2y - z)^{2 }= x^{2 }+ 4y^{2 }+ z^{2 }+ 4xy - 4yz - 2xz

So, the expansion of (x + 2y - z)^{2 }is

x^{2 }+ 4y^{2 }+ z^{2 }+ 4xy - 4yz - 2xz

To get formula / expansion for (a - b + c)^{2}, let us consider the formula / expansion for (a + b + c)^{2}.

The formula or expansion for (a + b + c)^{2 }is

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca**

In (a + b + c)^{2}, if b is negative, then we have

(a - b + c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find "b".

They are b^{2}, ab, bc.

Even if we take negative sign for "b" in b^{2}, the sign of b^{2} will be positive. Because it has even power 2.

The terms ab, bc will be negative. Because both "a" and "c" are multiplied by "b" that is negative.

Finally, we have

**(a - b + c) ^{2} = a^{2} + b^{2} + c^{2} - 2ab - 2bc + 2ac**

**Example :**

Expand :

(3x - y + 2z)^{2}

**Solution : **

(3x - y + 2z)^{2 }is in the form of (a - b + c)^{2}

Comparing (a + b - c)^{2} and (3x - y + 2z)^{2}, we get

a = 3x

b = y

c = 2z

Write the formula / expansion for (a - b + c)^{2}.

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

Substitute 3x for a, y for b and 2z for c.

(3x - y + 2z)^{2 }:

= (3x)^{2 }+ y^{2 }+ (2z)^{2 }- 2(3x)(y) - 2(y)(2z) + 2(3x)(2z)

(3x - y + 2z)^{2 }= 9x^{2 }+ y^{2 }+ 4z^{2 }- 6xy - 4yz + 12xz

So, the expansion of (3x - y + 2z)^{2}^{ }is

9x^{2 }+ y^{2 }+ 4z^{2 }- 6xy - 4yz + 12xz

To get the formula / expansion for (a - b - c)^{2}, let us consider the formula / expansion for (a + b + c)^{2}.

The formula or expansion for (a + b + c)^{2 }is

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca**

In (a + b + c)^{2}, if b and c are negative, then we have

(a - b - c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find "b" and "c".

They are b^{2}, c^{2}, ab, bc, ac.

Even if we take negative sign for "b" in b^{2} and negative sign for "c" in c^{2}, the sign of both b^{2 }and c^{2} will be positive. Because they have even power 2.

The terms "ab" and "ac" will be negative.

Because, in "ab", "a" is multiplied by "b" that is negative.

Because, in "ac", "a" is multiplied by "c" that is negative.

The term "bc" will be positive.

Because, in "bc", both "b" and "c" are negative.

That is,

negative ⋅ negative = positive

Finally, we have

**(a - b - c) ^{2} = a^{2} + b^{2} + c^{2} - 2ab + 2bc - 2ac**

**Example :**

Expand :

(x - 2y - 3z)^{2}

**Solution : **

(x - 2y - 3z)^{2 }is in the form of (a - b - c)^{2}

Comparing (a - b - c)^{2} and (x - 2y - 3z)^{2}, we get

a = x

b = 2y

c = 3z

Write the formula / expansion for (a - b - c)^{2}.

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

Substitute x for a, 2y for b and 3z for c.

(x - 2y - 3z)^{2 }:

= x^{2 }+ (2y)^{2 }+ (3z)^{2 }- 2(x)(2y) + 2(2y)(3z) - 2(x)(3z)

(x - 2y - 3z)^{2 }= x^{2 }+ 4y^{2 }+ 9z^{2 }- 4xy + 12yz - 6xz

So, the expansion of (x - 2y - 3z)^{2}^{ }is

x^{2 }+ 4y^{2 }+ 9z^{2 }- 4xy + 12yz - 6xz

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