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Mathematics

Assertion (A): ab+bc+ca−a2−b2−c23abc−a3−b3−c3\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}

= 1a+b+c\dfrac{1}{a + b + c}

Reason (R): ab+bc+ca−a2−b2−c23abc−a3−b3−c3\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}

= a2+b2+c2−ab−bc−caa3+b3+c3−3abc\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{a^3 + b^3 + c^3 - 3abc}

= a2+b2+c2−ab−bc−ca(a+b+c)(a2+b2+c2−ab−bc−ca)\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Expansions

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Answer

We know,

a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

So, 3abc − a3 − b3 − c3 = −(a3 + b3+ c3 −3abc)

= −(a + b + c)(a2 + b2+ c2 − ab − bc − ca)

Given, ab+bc+ca−a2−b2−c23abc−a3−b3−c3\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}

= −(a2+b2+c2−ab−bc−ca)−(a+b+c)(a2+b2+c2−ab−bc−ca)\dfrac{-(a^2 + b^2 + c^2 − ab − bc − ca )}{−(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}

= (a2+b2+c2−ab−bc−ca)(a+b+c)(a2+b2+c2−ab−bc−ca)\dfrac{(a^2 + b^2 + c^2 − ab − bc − ca )}{(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}

= 1a+b+c\dfrac{1}{a + b + c}

So, assertion (A) is true.

By formula,

a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

Given expression, ab+bc+ca−a2−b2−c23abc−a3−b3−c3\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}

= a2+b2+c2−ab−bc−caa3+b3+c3−3abc\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{a^3 + b^3 + c^3 - 3abc}

= a2+b2+c2−ab−bc−ca(a+b+c)(a2+b2+c2−ab−bc−ca)\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}

So, Reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

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