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jok3333 [9.3K]
2 years ago
14

How would the expression x^3+64 be rewritten using sum off cubes

Mathematics
2 answers:
Mamont248 [21]2 years ago
8 0
\bf \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
\boxed{64=4^3}\qquad x^3+64\implies x^3+4^3\implies (x+4)(x^2-x4+4^2)
\\\\\\
(x+4)(x^2-4x+16)
gogolik [260]2 years ago
3 0

Answer:

(x+4)(x^2-4x+4^2)

Step-by-step explanation:

We have been given an expression x^3+64 and we are asked to rewrite our expression using sum of cubes.

Sum of cubes: a^3+b^3=(a+b)(a^2-ab+b^2).

We can rewrite 64 as: 64=(4*4*4)=4^3

This means that a = x and b = 4, Upon substituting these values in sum of cubes formula we will get,

x^3+64=(x+4)(x^2-4x+4^2)

x^3+64=(x+4)(x^2-4x+4^2)

Therefore, after rewriting our given expression as sum of cubes we will get: (x+4)(x^2-4x+4^2).

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Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is Q(x,y) is predicate that x then x^{2}. where x,y are rational numbers.

a)

when x=-2, y=1

Here -2 that is x  satisfied. Then

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4 this is wrong. since 4>1

That is x^{2}>y^{2} Thus Q(x,y) =Q(-2,1)is false.

b)

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c)

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d)

Assume Q(x,y)=Q(9,10)

Here 9 satisfies the condition x

Then 9^{2}

81 satisfies the condition x^{2}.

Thus, Q(9,10) point exists and it is true. This satisfies the same values as in part (c)

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