# Expansion of (a ± b ± c)$$^{2}$$

We will discuss here about the expansion of (a ± b ± c)$$^{2}$$.

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

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

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

= sum of squares of a, b, c + 2(sum of the products of a, b, c taking two at a time}.

Therefore, (a – b + c)$$^{2}$$ = a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ + 2(ac – ab – bc)

Similarly for (a – b – c)$$^{2}$$, etc.

Corollaries:

(i) a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ = (a + b + c)$$^{2}$$ – 2(ab + bc + ca)

(ii) ab + bc + ca = $$\frac{1}{2}$${(a + b + c)$$^{2}$$ – (a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$)}

Solved Examples on Expansion of (a ± b ± c)$$^{2}$$

1. Expand (2x + y +3z)^2

Solution:

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

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

= 4x$$^{2}$$ + y$$^{2}$$ + 9z$$^{2}$$ + 4xy + 6yz + 12zx.

2. Expand (a - b - c)$$^{2}$$

Solution:

(a - b - c)$$^{2}$$

= a$$^{2}$$ + (-b)$$^{2}$$ + (-c)$$^{2}$$ + 2{a ∙ (-b) + (-b) ∙ (-c) + (-c) ∙ a}

= a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ - 2ab + 2bc - 2ca.

3. Expand (m - $$\frac{1}{2x}$$ + m$$^{2}$$)$$^{2}$$

Solution:

(m - $$\frac{1}{2x}$$ + m$$^{2}$$)$$^{2}$$

m$$^{2}$$ + (-$$\frac{1}{2m}$$)$$^{2}$$ + (m$$^{2}$$)$$^{2}$$ + 2{m ∙ (-$$\frac{1}{2m}$$) + (-$$\frac{1}{2m}$$) ∙ m$$^{2}$$  + m$$^{2}$$  ∙ m}

= m$$^{2}$$ + $$\frac{1}{4m^{2}}$$+ m$$^{4}$$ + 2{-$$\frac{1}{2}$$ - $$\frac{1}{2}$$m + m$$^{3}$$}

= m$$^{2}$$ + $$\frac{1}{4m^{2}}$$+ m$$^{4}$$ - 1 - m + 2m$$^{3}$$.

4. If p + q + r = 8 and pq + qr + rp = 18, find the value of p$$^{2}$$ + q$$^{2}$$ + r$$^{2}$$.

Solution:

We know that p$$^{2}$$ + q$$^{2}$$ + r$$^{2}$$ = (p + q + r)$$^{2}$$ - 2(pq + qr + rp).

Therefore, p$$^{2}$$ + q$$^{2}$$ + r$$^{2}$$

= 8$$^{2}$$ - 2 × 18

= 64 – 36

= 28.

5. If x – y – z = 5 and x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ = 29, find the value of xy – yz – zx.

Solution:

We know that ab + bc + ca = $$\frac{1}{2}$$[(a + b + c)$$^{2}$$ – (a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$)].

Therefore, xy + y(-z) + (-z)x = $$\frac{1}{2}$$[(x + y - z)$$^{2}$$ – (x$$^{2}$$ + y$$^{2}$$ + (-z)$$^{2}$$)]

Or, xy – yz – zx = $$\frac{1}{2}$$[5$$^{2}$$ – (x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$)]

= $$\frac{1}{2}$$[25 – 29]

= $$\frac{1}{2}$$(-4)

= -2.

9th Grade Math

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