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Y_Kistochka [10]
3 years ago
5

What is the factorization of 729^15+1000?

Mathematics
1 answer:
igomit [66]3 years ago
8 0

Answer:

The factorization of 729x^{15} +1000 is (9x^{5} +10)(81x^{10} -90x^{5} +100)

Step-by-step explanation:

This is a case of factorization by <em>sum and difference of cubes</em>, this type of factorization applies only in binomials of the form (a^{3} +b^{3} ) or (a^{3} -b^{3}). It is easy to recognize because the coefficients of the terms are <u><em>perfect cube numbers</em></u> (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of 729x^{15} +1000 by using the <em>sum and difference of cubes </em>factorization.

1.) We calculate the cubic root of each term in the equation 729x^{15} +1000, and the exponent of the letter x is divided by 3.

\sqrt[3]{729x^{15}} =9x^{5}

1000=10^{3} then \sqrt[3]{10^{3}} =10

So, we got that

729x^{15} +1000=(9x^{5})^{3} + (10)^{3} which has the form of (a^{3} +b^{3} ) which means is a <em>sum of cubes.</em>

<em>Sum of cubes</em>

(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})

with a= 9x^{5} y b=10

2.) Solving the sum of cubes.

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)

.

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