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Константин171
@Константин171
July 2022
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Разложите многочлены на множители
а) (2с + 1)³ - 64
б) p³ + (3p - 4)³
в) 8 - (3 - k)³
г) (5a + 4)³ - a³
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zinaidazina
Используем формулы
суммы кубов: а³ + b³ =
(a+b)(a²-ab+b²)
и
разности кубов: a³ - b³ =
(a-b)(a²+ab+b²)
а)
(2с + 1)³ - 64 = (2c+1)
³ - 4³ =
= (2c+1-4)·((2c+1)²+(2c+1)·4+4²) =
= (2c-3)(4c²+4c+1+8c+4+16) =
= (2c-3)·(4c²+12c+21)
б)
p³ + (3p - 4)³ =
= (p
+ (3p-4))·(p²- p·(3p-4)+(3p-4)²) =
= (4p-4)·(p²-3p²+4p+9p²-24p+16) =
= 4·(p-1)·(7p²-20p+16)
в)
8 - (3 - k)³ = 2
³ - (3-k)³ =
= (2- (3-k))·(2²+2·(3-k)+(3-k)²) =
= (-1+k)·(4+6-2k+9-6k+k²) =
= (k-1)·(k²-8k+19)
г)
(5a + 4)³ - a³ =
= (5a+4-a)·((5a+4)²+(5a+4)·a+a²) =
= (4a+4)(25a²+40a+16+5a²+4a+a²) =
= 4·(a+1)·(31a²+44a+16)
3 votes
Thanks 3
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Answers & Comments
суммы кубов: а³ + b³ = (a+b)(a²-ab+b²)
и
разности кубов: a³ - b³ = (a-b)(a²+ab+b²)
а)
(2с + 1)³ - 64 = (2c+1)³ - 4³ =
= (2c+1-4)·((2c+1)²+(2c+1)·4+4²) =
= (2c-3)(4c²+4c+1+8c+4+16) =
= (2c-3)·(4c²+12c+21)
б)
p³ + (3p - 4)³ =
= (p + (3p-4))·(p²- p·(3p-4)+(3p-4)²) =
= (4p-4)·(p²-3p²+4p+9p²-24p+16) =
= 4·(p-1)·(7p²-20p+16)
в)
8 - (3 - k)³ = 2³ - (3-k)³ =
= (2- (3-k))·(2²+2·(3-k)+(3-k)²) =
= (-1+k)·(4+6-2k+9-6k+k²) =
= (k-1)·(k²-8k+19)
г)
(5a + 4)³ - a³ =
= (5a+4-a)·((5a+4)²+(5a+4)·a+a²) =
= (4a+4)(25a²+40a+16+5a²+4a+a²) =
= 4·(a+1)·(31a²+44a+16)