assume x^2+y^2=1. Show that (1-3x^2y^2)=x^6+y^6

Question

Assume that x²+y²=1

show that (1-3x²y²)=x⁶+y⁶

Answer 1

Likely the intent of this problem was to practice using sum of cubes and perfect square trinomial techniques.

Remember:

a³ + b³ = (a + b) (a²-ab+b²)

And:

a²+2ab+b² = (a + b)²

For the given problem let's start with x⁶ + y⁶ and use the above to transform into 1 - 3x²y².

x⁶ + y⁶

(x² + y²)(x⁴ - x²y² + y⁴) -- factoring a sum of cubes

x⁴ + y⁴ - x²y² --replace the x² + y² factor with 1 (given), and reorder the terms of the trinomial

x⁴ + 2x²y² + y⁴ - x²y² - 2x²y² -- add and subtract a term to make the 1st three terms a perfect square tri

(x² + y²)² - 3x²y² --rewrite the perfect square trinomial as binomial squared: combine similar terms

1 - 3x²y² -- replace x² + y² with 1 (given)

Q.E.D.

Answer 2

x^2 + y^2 = 1

show 1-3x^2y^2 = x^6+ y^6

graph the equations, separately

surprise is they are both the same graph, the same circle with center at the origin and radius = 1

as USMC Gomer Pyle often said to marine sergeant Carter "surprise, surprise"

x^6 +y^6 = 1-3x^2y^2 use sum of cubes formula

(x^2+y^2)(x^4 -x^2y^2+ y^4) = 1-3x^2y^2

x^4 -x^2y^2 +y^4 = 1-3x^2y^2

x^4 +2x^2y^2 +y^4 = 1

(x^2+y^2)^2 = 1

1^2 = 1

1 =1

Answer 3

and the question is interchangeable with:

show cos⁶θ+sin⁶θ=1-(3/4)sin²2θ

Answer 4

See the videa.

https://youtu.be/6PEAvLO5zt4

Answer 5

We start with the first equation as it is less complicated and solve for either x^2 or y^2 to start. Lets solve for x^2 first. x^2+y^2=1 becomes x^2=1-y^2 after subtracting y^2 from both sides. First, notice that x^6=(x^2)^3 from our exponent rules. Starting with the right hand side of the second equation using x^2=1-y^2 as our definition for x^2, we substitute and get right hand side (RHS) = (1-y^2)^3+y^6. Carefully expanding the first term gives RHS=1-3y^2+3y^4-y^6+y^6=1-3y^2+3y^4=1-3y^2(1-y^2). Since x^2=1-y^2, substitution yields RHS=1-3(y^2)(x^2)=1-3(x^2)(y^2) which matches the left hand side of the second equation.

Answer 6

Take left side to begin with:

1-3x²y²

Substitute y²=1-x²

^{You will get an expression in x}

^{Do the same with the right side}

^{You'll get an expression in x that matches with what you get on the left side}

In other words, both sides will yield 1-3x^2+3x^4 after you substitute y²=1-x²

^{Hope this helps!}