Maths 4 Marks Questions

We have:
a³(b - c)³ + b³(c - a)³ + c³(a - b)³ =
[a(b - c)]³ + [a(b - c)]³ + [b(c - a)]³ + [c(a - b)]³
Put,
a(b - c) = x, b(c - a) = y, c(a - b) = z
Here,
x + y + z = a(b - c) + b(c - a) + c(a - b)
= ab - ac + bc - ab - ab + ac - bc
Thus,
We have:
a³(b - c)³ + b³(c - a)³ + c³(a - b)³ = x³ + y³ + z³
= 3xyz [When x + y + z = 0, x³ + y³ + z³ = 3xyz]
= 3a(b - c)b(c - a)c(a -b)
= 3abc(a - b)(b - c)(c - a)

(a + b + c)³ = [(a + b + c)]³ = (a + b)³ + c³ + 3(a + b)c(a + b + c)
⇒ (a + b + c)³ = a³ + b³ + 3ab(a + b) + c³ + 3(a + b)c(a + b + c)
⇒ (a + b + c)³ - a³ + b³ - c³ = 3ab(a + b) + 3(a + b)c(a + b + c)
⇒ (a + b + c)³ - a³ + b³ - c³ = 3(a + b)[ab + ca + cb + c²]
⇒ (a + b + c)³ - a³ + b³ - c³ = 3(a + b)[a(b + c) + c(b + c)]
⇒ (a + b + c)³ - a³ + b³ - c³ = 3(a + b)(b + c)(a +c)

(-12)³ + 7³ + 5³
We know:
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
x³ + y³ + z³ = (x + y + z)(x² + y² + z² - xy - yz - zx) + 3xyz
Here, x = (-12), y = 7, z = 5
(-12)³ + 7³ + 5³
= (-12 + 7 + 5)[(-12)² + 7² + 5² - 7(-12) - 35 + 60] + 3(-12) × 35
= 0 - 1260
= -1260

(28)³ + (-15)³ + (-13)³
We know:
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
x³ + y³ + z³ = (x + y + z)(x² + y² + z² - xy - yz - zx) + 3xyz
Here, x = (-28), y = -15, z = -13
(28)³ + (-15)³ + (-13)³
= (28 - 15 - 13)[(28)² + (-15)² + (-13)² - 28(-15) - (-15)(-13) - 28(-13)] + 3 × 28(-15)(-13)
= 0 + 16380
= 16380