Some Questions

1.
(a + b - c)(ab - bc - ca) + abc
= a²b - abc - ca² + ab² - b²c - abc - abc +bc² + c²a + abc
= a²b - ca² + ab² - b²c + bc² + c²a - 2abc
= (a²b - abc) + (c²a - ca²) + (ab² - abc) + (bc² - b²c)
= [ab(a - c) - ca(a - c)] + [ab(b - c) - bc(b - c)]
= (ab - ca)(a - c) + (ab - bc)(b - c)
= a(b - c)(a - c) + b(a - c)(b - c)
= (a + b)(b - c)(a - c)


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2.
(k + 1)x - (2k + 3)y - 3k - 5 = 0
(k + 1)x - (2k + 3)y = 3k + 5
kx + x - 2ky - 3y = 3k + 5
(x - 2y)k + (x - 3y) = 3k + 5

k can be any value if :
x - 2y = 3 ...... [1]
x - 3y = 5 ...... [2]

[1] - [2] :
y = -2

Put y = -2 into [1] :
x - 2(-2) = 3
x + 4 = 3
x = -1

Hence, when x = -1 and y = -2, the equation are true forany value of k.


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3.
L.S.
= (a + b)(b + c)(c + a) + abc
= (a + b)(bc + ab + c² + ca) + abc
= a(bc + ab + c² + ca) + b(bc + ab + c² + ca) + abc
= abc + a²b + ac² + ca² + b²c + ab² + bc² + abc + abc
= (a²b + ab² + abc) + (abc + b²c + bc²) + (a²c + abc + ac²)
= ab(a + b + c) + bc(a + b+ c) + ac(a + b + c)
= (ab + bc + ca)(a + b + c)
= (ab + bc + ca)(0)
= 0
= R.S.

Hence, (a + b)(b + c)(c + a) + abc = 0


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4.
L.S.
= (a + b + c)(ab + bc + ca)
= a²b + abc + ca² + ab² + b²c + abc + abc + bc² +c²a
= (abc + a²b + ac² + ca²) + (b²c + ab² + bc² + abc) + abc
= a(bc + ab + c² + ca) + b(bc + ab + c² + ca) + abc
= (a + b)(bc + ab + c² + ca) + abc
= (a + b)[b(c + a) + c(c + a)] + abc
= (a + b)(b + c)(c + a) + abc
= R.S.

Hence, (a + b + c)(ab + bc + ca) = (a + b)(b + c)(c + a) + abc