☞ (a+b)²= (a-b)²+4ab
☞ (a-b)²= a²-2ab+b²
☞ (a-b)²= (a+b)²-4ab
☞ a² + b²= (a+b)²-2ab.
☞ a² + b²= (a-b)²+2ab.
☞ a²-b²= (a +b)(a -b)
☞ 2(a²+b²)= (a+b)²+(a-b)²
☞ 4ab = (a+b)²-(a-b)²
☞ ab = {(a+b)/2}²-{(a-b)/2}²
☞ (a+b+c)² = a²+b²+c²+2(ab+bc+ca)
☞ (a+b)³ = a³+3a²b+3ab²+b³
☞ (a+b)³ = a³+b³+3ab(a+b)
☞ a-b)³= a³-3a²b+3ab²-b³
☞ (a-b)³= a³-b³-3ab(a-b)
☞ a³+b³= (a+b) (a²-ab+b²)
☞ a³+b³= (a+b)³-3ab(a+b)
☞ a³-b³ = (a-b) (a²+ab+b²)
☞ a³-b³ = (a-b)³+3ab(a-b)
☞ (a2 + b2 + c2) = (a + b + c) 2 – 2(ab + bc +
ca)
☞ 2 (ab + bc + ca) = (a + b + c) 2 – (a2 + b2 +
c2)
☞ (a + b + c) 3 = a3 + b3 + c3 + 3 (a + b) (b + c)
(c + a)
☞ a3 + b3 + c3 – 3abc = (a+b+c)(a2 + b2+ c2–
ab–bc– ca)
☞ a3 + b3 + c3 – 3abc = (a+b+c) { (a–b) 2+(b–c)
2+(c–a) 2 }
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☞ (x + a) (x + b) = x2 + (a + b) x + ab
☞ (x + a) (x – b) = x2 + (a – b) x – ab
☞ (x – a) (x + b) = x2 + (b – a) x – ab
☞ (x – a) (x – b) = x2 – (a + b) x + ab
☞ (x+p) (x+q) (x+r) = x3 + (p+q+r) x2 + (pq+qr
+rp) x +pq.
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