Use of identity for sum of cubes:
- x³ + y³ = (x + y)(x² - xy + y²)
Change it as:
- x³ + y³ = (x + y)(x² + 2xy + y² - 3xy) = (x + y)[(x + y)² - 3xy)]
It is easy to notice that:
- 64a³ = (4a)³ and
- 125b³ = (5b)³.
So the sum of cubes for the given expression will be:
- 64a³ + 125b³ =
- (4a + 5b)[(4a + 5b)² - 3(4a*5b)] =
- (4a + 5b)[(4a + 5b)² - 60ab]
Now substitute the values and calculate each of these:
(a) 4a + 5b = 5, ab = -1.
- (4a + 5b)[(4a + 5b)² - 60ab] =
- 5[5² - 60(-1)] =
- 5(25 + 60) =
- 5(85) =
- 425
(b) 4a + 5b = -2, ab = 1/15.
- (4a + 5b)[(4a + 5b)² - 60ab] =
- -2[(-2)² - 60(1/15)] =
- -2(4 - 4) =
- -2(0) =
- 0
(c) 4a + 5b = 1/3, ab = -1/9.
- (4a + 5b)[(4a + 5b)² - 60ab] =
- (1/3)[(1/3)² - 60(-1/9)] =
- (1/3)(1/9 + 60/9) =
- (1/3)(61/9) =
- 61/27 =
- 2 7/27
(d) 4a + 5b = -1, ab = 1.
- (4a + 5b)[(4a + 5b)² - 60ab] =
- -1[(-1)² - 60(1)] =
- -1(1 - 60) =
- -1(-59) =
- 59
