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factorization of 8a³ + 27b³ + 64c³ - 72 abc ??

As the first three terms seem to be cubes of 2a, 3b, 4c respectively, we can use the formula

(x+y+z)³ = x³ + y³ + z³ + 3x² (y+z) + 3y² (z+x) + 3z² (x+y) + 6xyz

So x³ + y³ + z³ = (x+y+z)³ - [ 3x² (y+z) + 3y² (z+x) + 3z² (x+y) + 6xyz ]

8a³ + 27b³ + 64c³

= (2a+3b+4c)³ - [ 3 4a² (3b+4c) + 3 9b² (4c+2a) + 3 16c² (2a+3b) + 6 2a 3b 4c ]

= (2a+3b+4c)³ - [ 12a² (3b+4c) + 27b² (4c+ 2a) + 48c² (2a+3b) + 144 abc ]

8a³ + 27b³ + 64c³ - 72 abc =

=(2a+3b+4c)³ - [36a²b+48a²c+108b²c+54b²a+96c²a+144c²b+144abc]-72abc

=(2a+3b+4c)³-[(36a²b+48a²c+72abc)+(108b²c+54b²a+72abc)+(96c²a+144c²b+72abc)]

= (2a+3b+4c)³-[12a(3ab+4ac+6bc)+18b(6bc+3ba+4ac)+24c(4ca+6cb+3ab)]

= (2a+3b+4c)³- [12a+18b+24c] [3ab+4ac+6bc]

= (2a+3b+4c) [ (2a+3b +4c)² - 6(3ab+4ac+6bc) ]

= (2a+3b+4c) [ 4a²+9b²+16c²+12ab+24bc+16ac -18ab -24ac -36bc]

As the first three terms seem to be cubes of 2a, 3b, 4c respectively, we can use the formula

(x+y+z)³ = x³ + y³ + z³ + 3x² (y+z) + 3y² (z+x) + 3z² (x+y) + 6xyz

So x³ + y³ + z³ = (x+y+z)³ - [ 3x² (y+z) + 3y² (z+x) + 3z² (x+y) + 6xyz ]

8a³ + 27b³ + 64c³

= (2a+3b+4c)³ - [ 3 4a² (3b+4c) + 3 9b² (4c+2a) + 3 16c² (2a+3b) + 6 2a 3b 4c ]

= (2a+3b+4c)³ - [ 12a² (3b+4c) + 27b² (4c+ 2a) + 48c² (2a+3b) + 144 abc ]

8a³ + 27b³ + 64c³ - 72 abc =

=(2a+3b+4c)³ - [36a²b+48a²c+108b²c+54b²a+96c²a+144c²b+144abc]-72abc

=(2a+3b+4c)³-[(36a²b+48a²c+72abc)+(108b²c+54b²a+72abc)+(96c²a+144c²b+72abc)]

= (2a+3b+4c)³-[12a(3ab+4ac+6bc)+18b(6bc+3ba+4ac)+24c(4ca+6cb+3ab)]

= (2a+3b+4c)³- [12a+18b+24c] [3ab+4ac+6bc]

= (2a+3b+4c) [ (2a+3b +4c)² - 6(3ab+4ac+6bc) ]

= (2a+3b+4c) [ 4a²+9b²+16c²+12ab+24bc+16ac -18ab -24ac -36bc]

__= (2a+3b+4c) [ 4a²+9b²+16c²-6ab -12bc - 8ac]__