Show how you obtain the product of 3 numbers if their sum is 10, the sum of their squares 38, and the sum of their cubes 160?
To summarize your question, the following equations are what we have[math]a + b + c = 10[/math][math]a^2 + b^2 + c^2 = 38[/math][math]a^3 + b^3 + c^3 = 160[/math]First, let’s have a look at something called the Elementary symmetric polynomial (ESP). In this case with 3 variables [math]a, b, c[/math], the corresponding polynomials are[math]p = a + b + c[/math][math]q = ab + bc + ca[/math][math]r = abc[/math]We are interested in those special polynomials because once the values of them are known, each of [math]a, b, c[/math] is easily computed due to Vieta’s Theorem. Besides, one more useful characteristic is that every symmetric expressions could be represented by those ESPs, which means, given a sufficient number of symmetric equations, we can always obtain the values of those ESPs.Back to our specific problem, we will try to express the sum of square and the sum of cube by [math]p, q, r[/math], as denoted above. As our concern is to find [math]r[/math], there is no need to compute the exact value of [math]a, b, c[/math].[math]p = a + b + c = 10[/math][math]a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = p^2 - 2q = 38[/math][math]a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + bc + ca) + 3abc = p^3 - 3pq + 3r = 160[/math]It is simple enough to solve the above system of equations of [math]p, q, r[/math]. The value of [math]r = abc[/math] to be obtained is [math]30[/math].Please note that there might be several triplets of [math](a,b,c)[/math] that satisfy the conditions. Their product, however, holds the consistent value of [math]30[/math] as computed.
How do I prove that three points are collinear?
Based on my long expirement with Maths,Here are some common ways,First method:Use the concept, if ABC is a straight line than, AB+BC=ACSecond method :In case of geometry, if you are given 3 ponits, A(x,y,z) ,B(a,b,c),C(p,q,r)Find the distance between AB =√(x-a)^2 + (y-b)^2 + (z-c)^2, then find BC and AC in similar way.If AB+BC=AC then points are collinear.Third method:Use the concept that area of the triangle formed by three collinear is zero.One way is by Using determinant,The other way is, Let A,B,C be there points, using coordinates, make two vector a (vector)=AB and b (vector) =BCNow a×b=0 (i.e a vector cross b vector=0)Forth meathod:If direction ratios of three vectors a,b,c are proportional then they are collinear.Thankyou!!
How do I solve for x,y, z when given x+y+z=6, xy+yz+zx=15 and xyz=14?
By adding all the three equations.x+y+z+xy+yz+zx+xyz+1=6+15+14+1=36=> (x+1)*(y+1)*(z+1) =36.= 1*2*18 = 2*3*6= 1*3*12 = -1*-2*18=……..=> By solving LHS with each of the RHS values , we get many non-negatve, negative and zero solutions of x,y,z.But as xyz not equal to zero => x,y,z not equal to zero.Select the x,y,z pairs to satisfy the above conditions.But here, in this it's time consuming.Method-2.X(y+z)+yz=15X(6-x) + 14/X = 156x-x^2+14/x =156x^2 - x^3 +14 =15x=> X^3 - 6x^2 +15x -14= 0Solving this gives 6 set of solutions.