Factor the binomial completely.
[tex]y^3-512[/tex]
Select the correct choice below and, if necessary
Answer (2)
The binomial y 3 − 512 can be factored using the difference of cubes formula, yielding ( y − 8 ) ( y 2 + 8 y + 64 ) . The quadratic factor y 2 + 8 y + 64 cannot be factored further due to its negative discriminant. Therefore, the complete factorization is ( y − 8 ) ( y 2 + 8 y + 64 ) .
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Recognize the binomial as a difference of cubes: y 3 − 512 = y 3 − 8 3 .
Apply the difference of cubes factorization formula: a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) .
Substitute a = y and b = 8 into the formula: ( y − 8 ) ( y 2 + 8 y + 64 ) .
Check if the quadratic factor can be factored further; since the discriminant is negative, it cannot be factored further. The final answer is ( y − 8 ) ( y 2 + 8 y + 64 ) .
Explanation
Understanding the Problem We are asked to factor the binomial y 3 − 512 completely. Notice that this is a difference of cubes. Our goal is to express this binomial as a product of simpler factors.
Recalling the Formula Recall the difference of cubes factorization formula: a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) . We need to identify a and b in our expression.
Identifying a and b In our case, we have y 3 − 512 . We can rewrite 512 as 8 3 since 8 × 8 × 8 = 512 . Thus, we have y 3 − 8 3 . Here, a = y and b = 8 .
Applying the Formula Now, substitute a = y and b = 8 into the difference of cubes formula:
y 3 − 8 3 = ( y − 8 ) ( y 2 + y × 8 + 8 2 ) = ( y − 8 ) ( y 2 + 8 y + 64 ) .
Checking the Quadratic Factor Next, we need to check if the quadratic factor y 2 + 8 y + 64 can be factored further. To do this, we can calculate the discriminant, which is given by D = b 2 − 4 a c , where a = 1 , b = 8 , and c = 64 in the quadratic a y 2 + b y + c .
D = 8 2 − 4 × 1 × 64 = 64 − 256 = − 192 .
Final Factorization Since the discriminant D = − 192 is negative, the quadratic factor y 2 + 8 y + 64 has no real roots and cannot be factored further using real numbers. Therefore, the complete factorization of y 3 − 512 is ( y − 8 ) ( y 2 + 8 y + 64 ) .
Examples
The difference of cubes factorization is useful in various engineering and physics applications. For example, consider designing a container that needs to hold a specific volume of liquid. If the volume can be expressed as a difference of cubes, factoring it can help determine the dimensions of the container. Also, in signal processing, certain signal transformations can be simplified using algebraic identities like the difference of cubes. Factoring helps in simplifying complex expressions and making them easier to analyze and implement.
