Question

if the roots of the equation(c²-ab)x²-2(a²-bc)x+(b²-ac)=0 are real and equal, show that either a=0 or (a³+b³+c³)=3abc.
please make it simple not complicated.....
Don't copy.........​

Answer 1

$\large\bf\underline{Given:-}$

• Equation:- (c² - ab)x² - 2(a² - bc)x + (b² - ac)

$\large\bf\underline {To \: find:-}$

• we need to show that either a = 0 or (a³ + b³ + c³ = 3abc.

$\huge\bf\underline{Solution:-}$

Given quadratic equation:-

• (c² - ab)x² - 2(a² - bc)x + (b² - ac)

where ,

• a = (c² - ab)
• b = -2(a² - bc)
• c = (b² - ac)

Roots of quadratic equation are equal so,

• D = 0
• ≫ D = b² - 4ac
• b² - 4ac = 0

» [-2(a² - bc)]² - 4 × (c² - ab) × (b² - ac)

» 4(a⁴ + b²c² - 2a²bc = 4[c²(b²-ac)-ab(b²-ac)]

» a⁴ + b²c² - 2a²bc = b²c² - ac³ - ab³ + a²bc

» a⁴ + b²c²- b²c² -2a²bc -a²bc = -ac³ - ab³

» a⁴ + ac³+ ab³ - 3a²bc = 0

» a(a³ + b³ + c³ - 3abc) = 0

So, either a = 0

• Or

» a³ + b³ + c³ - 3abc = 0

» a³ + b³ + c³ = 3abc

Hence, Proved

either a = 0 or a³ + b³ + c³ = 3abc.

$\rule{200}3$

Answer 2

Step-by-step explanation:

$\mathfrak{ \large \bold \red{ \underline{ \underline{question}}}} \\ \\ { \large{ ({c }^{2} - ab) {x}^{2} - 2( {a}^{2} - bc)x + ( {b}^{2} - ac) = 0 }} \\ \\ { \large{show \: that \: a = 0 \: or \: ( {a}^{3} + {b}^{3} + {c}^{3} ) = 3abc}} \\ \\ { \large \bold \green{ \underline{solution}}} \\ { \large{ \underline{a = ( {c}^{2} - ab) and \: b = - 2( {a}^{2} - bc) \: and \: c = ( {b}^{2} - ac) }}} \\ \\ { \large \orange{ \underline{since \: root \: are \: equal \: so \: d = 0}}} \\ { \large \blue{ \underline{formula \: \: d = {b}^{2} - 4ac }}} \\ { \large{( - 2( {a}^{2} - bc)) {}^{2} - 4( {c }^{2} - ab)( {b }^{2} - ac) = 0 }} \\ { \large{4( {a}^{4} - 2 {a}^{2}c + {b }^{2} {c}^{2} ) - 4( {b}^{2} {c }^{2} - a {c}^{3} - a {b}^{3} + {a}^{2} bc) = 0}} \\ { \large{ {a}^{4} - 3 {a}^{2} bc + a {c}^{3} + a {b}^{3} = 0 }} \\{ \large{a( {a}^{3} - 3abc + {a}^{3} + {b}^{3}) = 0 }} \\ { \large{a = 0 \: \: or \: {a}^{3} + {c}^{3} + {b}^{3} = 3abc}} \\ \\ { \large \bold \pink{ \underline{hence \: prove}}}$