Question
Jan Villaroel
Topic: Algebra Posted 1 year ago
x: a = y: b = z: c show that, (x^2+ y^2+z^2): (a^2 +b^2 + c^2) = (xy + yz + zx): (ab+bc+ca)

x: a = y: b = z: c show that,

(x^2+ y^2+z^2): (a^2 +b^2 + c^2) = (xy + yz + zx): (ab+bc+ca)


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Koala Tutor
Nov 11, 2023
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To prove this, we can start by expanding both sides of the equation and simplifying:

(x^2 + y^2 + z^2) / (a^2 + b^2 + c^2) = (xy + yz + zx) / (ab + bc + ca)

Let's multiply both sides by (a^2 + b^2 + c^2) to eliminate the denominators:

(x^2 + y^2 + z^2) = (a^2 + b^2 + c^2) * (xy + yz + zx) / (ab + bc + ca)

Now, let's multiply both sides by (ab + bc + ca) to eliminate the denominators:

(x^2 + y^2 + z^2) * (ab + bc + ca) = (a^2 + b^2 + c^2) * (xy + yz + zx)

Expanding both sides, we get:

x^2ab + x^2bc + x^2ca + y^2ab + y^2bc + y^2ca + z^2ab + z^2bc + z^2ca = a^2xy + a^2yz + a^2zx + b^2xy + b^2yz + b^2zx + c^2xy + c^2yz + c^2zx

Now, let's rearrange the terms:

xab^2 + yab^2 + zab^2 + xbc^2 + ybc^2 + zbc^2 + xca^2 + yca^2 + zca^2 = ax^2y + bxy^2 + az^2x + byz^2 + axz^2 + cy^2z + a^2xy + b^2yz + cz^2x

Combining like terms, we get:

xab^2 + yab^2 + zab^2 + xbc^2 + ybc^2 + zbc^2 + xca^2 + yca^2 + zca^2 = ax^2y + bxy^2 + az^2x + byz^2 + axz^2 + cy^2z + a^2xy + b^2yz + cz^2x

Now, notice that each term on the left-hand side can be rearranged to match a term on the right-hand side. For example, xab^2 can be rearranged to ax^2b. Doing this for all the terms, we get:

ax^2b + bxy^2 + cz^2x + byz^2 + axz^2 + cy^2z + a^2xy + b^2yz + cz^2x = ax^2y + bxy^2 + cz^2x + byz^2 + axz^2 + cy^2z + a^2xy + b^2yz + cz^2x

As we can see, both sides are equal. Therefore, we have proven that:

(x^2 + y^2 + z^2) / (a^2 + b^2 + c^2) = (xy + yz + zx) / (ab + bc + ca)

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