Phân tích thành nhân tử:
a) A = ab(a − b) + bc(b − c) + ca(c − a)
b) B = a(b2 − c2) + b(c2 − a2) + c(a2 − b2)
c) C = (a + b + c)3 − a3 − b3 − c3
Lời giải
a) A = ab(a − b) + bc(b − c) + ca(c − a)
= ab(a − b) + b2c − bc2 + c2a − a2c
= ab(a − b) + c2(a − b) − c(a2 − b2)
= ab(a − b) + c2(a − b) − c(a − b)(a + b)
= (a − b)[ab + c2 − c(a + b)]
= (a − b)(ab + c2 − ac − bc)
= (a − b)[a(b − c) − c(b − c)]
= (a − b)(b − c)(a − c).
b) B = a(b2 − c2) + b(c2 − a2) + c(a2 − b2)
= ab2 − ac2 + bc2 − a2b + c(a − b)(a + b)
= −ab(a − b) − c2(a − b) + c(a − b)(a + b)
= (a − b)[−ab − c2 + c(a + b)]
= (b − a)[ab + c2 − c(a + b)]
= (b − a)(ab + c2 − ac − bc)
= (b − a)[a(b − c) − c(b − c)]
= (b − a)(b − c)(a − c).
c) C = (a + b + c)3 − a3 − b3 − c3
= a3 + b3 + c3 + 3ab(a + b) + 3bc(b + c) + 3ca(c + a) + 6abc − a3 − b3 − c3
= 3ab(a + b) + 3bc(b + c) + 3ca(c + a) + 6abc
= 3(a2b + ab2 + a2c + ac2 + b2c + bc2 + 2abc)
= 3[ab(a + b) + bc(a + b) + c2(a + b) + ac(a + b)]
= 3(a + b)(ab + bc + c2 + ac)
= 3(a + b)[b(a + c) + c(a + c)]
= 3(a + b)(a + c)(b + c).