Question

If xa = yb=zc and a + b + c =0 then prove that, xyz = 1.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

If x=a(b-c), y=b(c-a), z=c(a b) then prove that (x/a) ^3+(y/b) ^3+(z/c) ^3 = 3xyz/abc?

we know

a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

we have,

x=a(b-c) or x/a=(b-c)

y=b(c-a) or y/b=(c-a)

z=c(a-b) or z/c=(a-b)

now,

(x/a)^3+(y/b)^3+(z/c)^3–3(x/a)(y/b)(z/c)

=(x/a+y/b+z/c){(x/a)^2+(y/b)^2+(z/c)^2-(x/a)(y/b)-(y/b)(z/c)-(z/c)(x/a)}

={(b-c)+(c-a)+(a-b)}{ same}

={b-c+c-a+a-b}{ same }

=0{ same }=0

so

(x/a)^3+(y/b)^3+(z/c)^3=3(x/a)(y/b)(z/c)

=3xyz/abc proved