What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64
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-1
1 answer:
Answer:
(3 + 4x)(9 - 12x + 16x²) = 0
(4m - 1)(16m² + 4m + 1) = 0
Step-by-step explanation:
Here we have to solve the sum of two cubes which is
Now, the equation is
⇒ 3³ + (4x)³ = 0
⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0
⇒ (3 + 4x)(9 - 12x + 16x²) = 0
So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0
Therefore, from the above two relation we can solve for x.
One root will be and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.
Again, we have to solve the difference of two cubes which is
Now, the equation is
⇒ (4m)³ - 1³ = 0
⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0
⇒ (4m - 1)(16m² + 4m + 1) = 0
So, (4m - 1) = 0 or (16m² + 4m + 1) = 0
Therefore, from the above two relation we can solve for m.
One root will be and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation. (Answer)
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1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
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The number 1052, written as a base 6 number is 4512
Given: 1052 toothpicks
To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number
First we note that,
This implies that exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.
Now, and
. This implies that
exceeds 1052 and thus there can be at most 4 groups of
.
Then,
So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.
Now, and
. This implies that
exceeds 188 and thus there can be at most 5 groups of
.
Then,
So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.
Now, and
. This implies that
exceeds 8 and thus there can be at most 1 group of
.
Then,
So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.
Now, and
. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.
This concludes the grouping.
Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
Then,
So, the number 1052, written as a base 6 number is 4512.
Learn more about change of base of numbers here: