Let then, x = ak, y = bk, z = ck. Given, Putting values of x, y, z in above equation and solving L.H.S., Solving, R.H.S. now, Since, L.H.S. = R.H.S. Hence, proved that, .

Mathematics

If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ , prove that
$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3.$

Ratio Proportion

9 Likes

Answer

Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k,$ then,

x = ak, y = bk, z = ck.

Given,

$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$

Putting values of x, y, z in above equation and solving L.H.S.,

$= \dfrac{3(ak)^3 - 5(bk)^3 + 4(ck)^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = \dfrac{3a^3k^3 - 5b^3k^3 + 4c^3k^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = k^3\Big(\dfrac{3a^3 - 5b^3 + 4c^3}{3a^3 - 5b^3 + 4c^3}\Big) \\[1em] = k^3.$

Solving, R.H.S. now, $= \Big(\dfrac{3(ak) - 5(bk) + 4(ck)}{3a - 5b + 4c}\Big)^3 \\[1em] = \Big[k\Big(\dfrac{3a - 5b + 4c}{3a - 5b + 4c}\Big)\Big]^3 \\[1em] = k^3.$

Since, L.H.S. = R.H.S. Hence, proved that,

$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$ .

Answered By

5 Likes

Mathematics

If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ , prove that
$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3.$

Ratio Proportion

Answer

Related Questions

If q is the mean proportional between p and r, prove that :
$p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big).$

If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$ , prove that each ratio is equal to
(i) $\sqrt{\dfrac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$
(ii) $\Big(\dfrac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}.$

If x : a = y : b, prove that,
$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

If $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c},$ prove that each ratio is equal to
$\dfrac{x + y + z}{a + b + c}.$

Mathematics

If xa=yb=zc\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}ax​=by​=cz​, prove that3x3−5y3+4z33a3−5b3+4c3=(3x−5y+4z3a−5b+4c)3.\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3.3a3−5b3+4c33x3−5y3+4z3​=(3a−5b+4c3x−5y+4z​)3.

Ratio Proportion

Answer

Related Questions

If q is the mean proportional between p and r, prove that :p2−3q2+r2=q4(1p2−3q2+1r2).p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big).p2−3q2+r2=q4(p21​−q23​+r21​).

If x : a = y : b, prove that,x4+a4x3+a3+y4+b4y3+b3=(x+y)4+(a+b)4(x+y)3+(a+b)3.\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.x3+a3x4+a4​+y3+b3y4+b4​=(x+y)3+(a+b)3(x+y)4+(a+b)4​.

If xb+c−a=yc+a−b=za+b−c,\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c},b+c−ax​=c+a−by​=a+b−cz​, prove that each ratio is equal tox+y+za+b+c.\dfrac{x + y + z}{a + b + c}.a+b+cx+y+z​.

If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ , prove that
$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3.$

If q is the mean proportional between p and r, prove that :
$p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big).$

If x : a = y : b, prove that,
$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

If $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c},$ prove that each ratio is equal to
$\dfrac{x + y + z}{a + b + c}.$