X exponentiated by two plus negative Y raised to the power two ;How did you make exponents on the computer?The base is X;
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Algebra Expand using the Binomial Theorem (3xy)^3 (3x − y)3 ( 3 x y) 3 Use the binomial expansion theorem to find each term The binomial theorem states (ab)n = n ∑ k=0nCk⋅(an−kbk) ( a b) n = ∑ k = 0 n n C k ⋅ ( a n k b k) 3 ∑ k=0 3!⋅(x)3−k ⋅(−y)k ∑ k = 0 3 Result A sum containing 2 terms;
Algebra Factor x^3y^3 x3 − y3 x 3 y 3 Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 abb2) a 3 b 3 = ( a b) ( a 2 a b b 2) where a = x a = x and b = y b = y (x−y)(x2 xyy2) ( x y) ( x 2 x y y 2)Expand (xy)^3 (x y)3 ( x y) 3 Use the Binomial Theorem x3 3x2y3xy2 y3 x 3 3 x 2 y 3 x y 2 y 3The base is Y;
In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial According to the theorem, it is possible to expand the polynomial n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positive integer depending on n and b For example,The second term of the sum is equal to a negative power; For expanding the equation we know the formula of cube of two number which is given below Now we will compare the given equation with the formula we will get the value of a and b which is Put these value in above equation and simplify it Now simplify the above expand form to get the answer Now further simplify the terms
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Answer The expansion of (xy) 3 is x 3 y 3 3x 2 y 3xy 2 Let us see how to expand (xy) 3 Explanation The expression (xy) 3 can be written as, (xy)(xy)(xy) First simplify (xy)(xy) by binomial multiplication (xy)(xy) = x 2 2xy y 2 Now multiply (xy) with x 2 2xy y 2 (xy)(xy)(xy) = (xy)(x 2 2xy y 2) = x 3 2x 2 y xy 2 yx 2 2xy 2 y 3 = x 3 3x 2 y 3xy 2 yAlgebra Expand using the Binomial Theorem (xy)^3 (x − y)3 ( x y) 3 Use the binomial expansion theorem to find each term The binomial theorem states (ab)n = n ∑ k=0nCk⋅(an−kbk) ( a b) n = ∑ k = 0 n n C k ⋅ ( a n k b k) 3 ∑ k=0 3!Expand 2(x y) 3(x y) Equations Basic (Linear) Solve For Quadratic Solve by Factoring Completing the Square Quadratic Formula Rational
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The exponent is two;The first term of the sum is a power;Expand using the Binomial Theorem (x3)^3 (x 3)3 ( x 3) 3 Use the binomial expansion theorem to find each term The binomial theorem states (ab)n = n ∑ k=0nCk⋅(an−kbk) ( a b) n = ∑ k = 0 n n C k ⋅ ( a n k b k) 3 ∑ k=0 3!
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It is clear that when $x=y$ we have $x^3y^3=0$ Then use long division to divide $x^3y^3$ by $xy$ and the result will be the equation on the right Another way would be to write $$\left(\frac{x}{y}\right)^3 1$$ Now we wish to find the zeros of this polynomialIn the case where x greatly exceeds y, the numerator is essentially 2 y ( 3 x) = 6 x y, and the denominator is approximately 2 x Therefore, the given expression is roughly 3 x 1 / 2 y, as claimedLearn about expand using our free math solver with stepbystep solutions
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Expand (xy)^3 full pad » x^2 x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot \msquare {\square} \le \ge (xy)3= (x² 2xy y²)(xy) Then you FOIL (First, outer, inner, last) (xy)3 = (x² 2xy y²)(xy) = x2x 2xxy xy2 x2y 2xyy y2y and when you combine like terms = x3 3x2y 3xy2 y3 Upvote •8Downvote Comments •4 More Report Tyler C Thank you, Kristin! Explanation note that (x a)3 = x3 (a a a)x2 (aa aa aa)x a3 (x −1)3 → a = −1 ⇒ (x − 1)3 = x3 ( −1 − 1 − 1)x2 (1 1 1)x ( −1)3 = x3 −3x2 3x − 1
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Expand (x2)^3 (x 2)3 ( x 2) 3 Use the Binomial Theorem x3 3x2 ⋅23x⋅ 22 23 x 3 3 x 2 ⋅ 2 3 x ⋅ 2 2 2 3 Simplify each term Tap for more steps Multiply 2 2 by 3 3 x 3 6 x 2 3 x ⋅ 2 2 2 3 x 3 6 x 2 3 x ⋅ 2 2 2 3 Raise 2 2 to the power of 2 2The exponent is two;
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