Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2
(ii) (2x − y + z)2
(iii) (−2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (–2x + 5y – 3z)2
(vi) ((1/4)a - (1/2)b + 1)2
(i) (x+2y+4z)2
Using identity, (x+y+z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)2 = x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x2+4y2+16z2+4xy+16yz+8xz
(ii) (2x−y+z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 2x
y = −y
z = z
(2x−y+z)2 = (2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x2+y2+z2–4xy–2yz+4xz
(iii) (−2x+3y+2z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = −2x
y = 3y
z = 2z
(−2x+3y+2z)2 = (−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x2+9y2+4z2–12xy+12yz–8xz
(iv) (3a –7b–c)2
Using identity (x + y + z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 3a
y = – 7b
z = – c
(3a –7b– c)2 = (3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a2 + 49b2 + c2– 42ab+14bc–6ca
(v) (–2x+5y–3z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = –2x
y = 5y
z = – 3z
(–2x+5y–3z)2 = (–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y×– 3z) + (2×–3z ×–2x)
= 4x2+25y2 +9z2– 20xy–30yz+12zx
(vi) ((1/4)a-(1/2)b+1)2
Using identity, (x + y + z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z = 1
((1/4)a - (1/2)b + 1)2 = \((\frac{1}{4}a)^2 + (- \frac{1}{2}b)^2 + (1)^2 + \) \((2 \times -\frac{1}{4}a \times -\frac{1}{2}b)\)+\((2 \times \frac{1}{2}b \times 1)\) + \((2 \times 1 \times \frac{1}{4}a)\)
= \(\frac{1}{16}a^2 \) + \(\frac{1}{4}b^2 + 1^2 - \frac{2}{8}ab - \frac{2}{2}b\) + \(\frac{2}{4}a\)
= \(\frac{1}{16}a^2 \) + \(\frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b\) + \(\frac{1}{2}a\)