answer:1. **Determine that ( P ) is an even function:** First, let us consider the given equation ( P(a - b) + P(b - c) + P(c - a) = 2P(a + b + c) ). By choosing specific values for ( a ), ( b ), and ( c ) such that ( a b + b c + c a = 0 ), we aim to understand the properties of ( P ). Let's take ( b = c = 0 ): [ a cdot 0 + 0 cdot 0 + 0 cdot a = 0 ] [ P(a - 0) + P(0 - 0) + P(0 - a) = 2P(a + 0 + 0) ] [ P(a) + P(0) + P(-a) = 2P(a) ] Since ( P(-a) = P(a) ), ( P ) must be an even function. Therefore, ( P ) only contains terms of even degree. 2. **Consider the degree and constant term of ( P ):** Because ( P ) is an even function, we can write: [ P(x) = a_n x^{2n} + a_{n-1} x^{2n-2} + cdots + a_1 x^2 + a_0 ] Evaluate at ( x = 0 ), we get: [ P(0-0) + P(0-0) + P(0-0) = 2P(0+0+0) ] [ P(0) + P(0) + P(0) = 2P(0) ] [ 3P(0) = 2P(0) ] This implies that: [ P(0) = 0 ] Therefore, the constant term ( a_0 ) must be zero. 3. **Analyzing specific values:** Next, try ( a = 6x ), ( b = 3x ), and ( c = -2x ): [ 6x cdot 3x + 3x cdot (-2x) + (-2x) cdot 6x = 18x^2 - 6x^2 - 12x^2 = 0 ] Substituting these values into the equation: [ P(6x - 3x) + P(3x - (-2x)) + P((-2x) - 6x) = 2P(6x + 3x - 2x) ] [ P(3x) + P(5x) + P(-8x) = 2P(7x) ] 4. **Comparing leading term coefficients:** Let the polynomial ( P(x) ) be of degree ( n ). The relationship above involves comparing the leading term (dominant term) coefficients: [ 3^n + 5^n + (-8)^n = 2 cdot 7^n ] Testing integer values of ( n ): 5. **Assess ( n geq 5 ):** For ( n = 5 ): [ 3^5 + 5^5 + (-8)^5 = 243 + 3125 - 32768 neq 2 cdot 16807 ] Clearly, this does not hold for ( n = 5 ) or greater. Similar evaluations will verify this for ( n > 5 ). 6. **Considering lower orders:** By reducing the degree, we limit ( P ) to: [ P(x) = a_n x^{4} + a_{2} x^{2} ] **Verification:** Substituting this form back into the initial equation ensures it meets the conditions specified. # Conclusion: Hence, the polynomial that meets the conditions is of the form: [ P(x) = a x^4 + b x^2 ] Thus, the polynomials that satisfy the given conditions are: [ %boxed{P(x) = ax^4 + bx^2} boxed{P(x) = ax^4 + bx^2} ]

answer:Define the line passing through begin{pmatrix} -6 3 end{pmatrix} and begin{pmatrix} 4 5 end{pmatrix} which can be parameterized as: [ begin{pmatrix} -6 3 end{pmatrix} + t left( begin{pmatrix} 4 5 end{pmatrix} - begin{pmatrix} -6 3 end{pmatrix} right) = begin{pmatrix} -6 3 end{pmatrix} + t begin{pmatrix} 10 2 end{pmatrix} = begin{pmatrix} 10t - 6 2t + 3 end{pmatrix}. ] The projection results should be same for mathbf{p}, including the vector begin{pmatrix} 1 2 end{pmatrix} on the same line: [ begin{pmatrix} 10t - 6 2t + 3 end{pmatrix} cdot begin{pmatrix} -10 2 end{pmatrix} = 0. ] Solving for t gives (10t - 6)(-10) + (2t + 3)(2) = 0 leading to -100t + 60 + 4t + 6 = 0 yielding t = frac{-66}{-96} = frac{33}{48}. Thus, [ mathbf{p} = boxed{begin{pmatrix} frac{10(33/48) - 6}{17/48 |} frac{2(33/48) + 3}{frac{17}{48}} end{pmatrix}} = boxed{begin{pmatrix} frac{2/3}{17/48} frac{69/48}{17/48} end{pmatrix}} = boxed{begin{pmatrix} frac{32}{17} frac{69}{17} end{pmatrix}}. ]

answer:To find the degree of the monomial -frac{2}{5}x^2y, we need to sum the exponents of all the variables in the monomial. The exponent of x is 2, and the exponent of y is 1. Therefore, the degree of the monomial is the sum of these exponents: [2 + 1 = 3] Thus, the degree of the monomial -frac{2}{5}x^2y is boxed{3}.

answer:1. Simplify the expression on the right side: sqrt[16]{4} = 4^{1/16} = (2^2)^{1/16} = 2^{1/8}. Thus, sqrt[16]{4} = 2^{1/8}. When this is multiplied by 64, we get: 64sqrt[16]{4} = 64 cdot 2^{1/8} = 2^6 cdot 2^{1/8} = 2^{49/8} . 2. Set up the equation: sqrt[4]{R^3} = 2^{49/8} . To solve for R, first square both sides to get rid of the fourth root: left(sqrt[4]{R^3}right)^4 = left(2^{49/8}right)^4 = 2^{49/2} = 2^{24.5} . Thus, R^3 = 2^{24.5}. 3. Now solve for R: R = sqrt[3]{2^{24.5}} = 2^{24.5/3} = 2^{8.1667} . 4. Since the exponent is fractional, this will simplify to: R = 2^{8 + 1/6} = 2^8 cdot 2^{1/6} = 256 cdot 2^{1/6} . Conclusion: The value of R is boxed{256 cdot 2^{1/6}}.