There are only two unknowns!?

Algebra Level 4

{ a x + b y = 8 a x 2 + b y 2 = 14 a x 3 + b y 3 = 62 a x 4 + b y 4 = 146 \large{ \begin{cases} ax + by = 8 \\ ax^2 + by^2 = 14 \\ ax^3 + by^3 = 62 \\ ax^4 + by^4= 146 \\ \end{cases}}

If a , b , x a,b,x and y y are real number satisfying the system of equations above, find a x 5 + b y 5 ax^5 + by^5 .


The answer is 518.

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1 solution

Choi Chakfung
Mar 14, 2016

a x + b y = 8 ( 1 ) a x 2 + b y 2 = 14 ( 2 ) a x 3 + b y 3 = 62 ( 3 ) a x 4 + b y 4 = 146 ( 4 ) f i n d a x 5 + b y 5 ( 4 ) × ( x + y ) ( a x 4 + b y 4 ) ( x + y ) = 146 ( x + y ) a x 5 + a x 4 y + b y 4 x + b y 5 = 146 ( x + y ) a x 5 + b y 5 + x y ( a x 3 + b y 3 ) = 146 ( x + y ) a x 5 + b y 5 + 62 x y = 146 ( x + y ) ( 5 ) ( 2 ) × ( x + y ) s i m i l a r l y a x 3 + b y 3 + 8 x y = 14 ( x + y ) 62 + 8 x y = 14 ( x + y ) ( 6 ) ( 3 ) × ( x + y ) s i m i l a r l y a x 4 + b y 4 + 14 x y = 62 ( x + y ) 146 + 14 x y = 62 ( x + y ) ( 7 ) ( 6 ) ÷ ( 7 ) 62 + 8 x y 146 + 14 x y = 14 62 s i m p l i f y i t x y = 6 ( 8 ) p u t x y = 6 i n t o ( 7 ) 146 14 × 6 = 62 ( x + y ) ( x + y ) = 1 ( 9 ) p u t ( 8 ) a n d ( 9 ) i n t o ( 5 ) a x 5 + b y 5 + 62 x y = 146 ( x + y ) a x 5 + b y 5 = 146 ( 1 ) 62 ( 6 ) a x 5 + b y 5 = 146 + 372 a x 5 + b y 5 = 518 ax+by=8\cdots (1)\\ a{ x }^{ 2 }+b{ y }^{ 2 }=14\cdots (2)\\ a{ x }^{ 3 }+b{ y }^{ 3 }=62\cdots (3)\\ a{ x }^{ 4 }+b{ y }^{ 4 }=146\cdots (4)\\ find\quad a{ x }^{ 5 }+b{ y }^{ 5 }\\ (4)\times (x+y)\\ (a{ x }^{ 4 }+b{ y }^{ 4 })(x+y)=146(x+y)\\ a{ x }^{ 5 }+a{ x }^{ 4 }y+b{ y }^{ 4 }x+b{ y }^{ 5 }=146(x+y)\\ a{ x }^{ 5 }+b{ y }^{ 5 }+xy(a{ x }^{ 3 }+b{ y }^{ 3 })=146(x+y)\\ a{ x }^{ 5 }+b{ y }^{ 5 }+62xy=146(x+y)\cdots (5)\\ (2)\times (x+y)\\ similarly\\ a{ x }^{ 3 }+b{ y }^{ 3 }+8xy=14(x+y)\\ 62+8xy=14(x+y)\cdots (6)\\ (3)\times (x+y)\\ similarly\\ a{ x }^{ 4 }+b{ y }^{ 4 }+14xy=62(x+y)\\ 146+14xy=62(x+y)\cdots (7)\\ (6)\div (7)\\ \frac { 62+8xy }{ 146+14xy } =\frac { 14 }{ 62 } \\ simplify\quad it\quad \\ xy=-6\cdots (8)\\ put\quad xy=-6\quad into\quad (7)\\ 146-14\times 6=62(x+y)\\ (x+y)=1\cdots (9)\\ put\quad (8)\quad and\quad \quad (9)\quad into\quad (5)\\ a{ x }^{ 5 }+b{ y }^{ 5 }+62xy=146(x+y)\\ a{ x }^{ 5 }+b{ y }^{ 5 }=146(1)-62(-6)\\ a{ x }^{ 5 }+b{ y }^{ 5 }=146+372\\ a{ x }^{ 5 }+b{ y }^{ 5 }=518

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