Question-211393

Question Number 211393 by BaliramKumar last updated on 08/Sep/24

Commented by BaliramKumar last updated on 08/Sep/24

a^{2} - b^{2} = 25 a + b = 5

(a - b) (a + b) = 25

(a - b) 5 = 25

a - b = 5 \dots \dots \dots (i)

a + b = 5 \dots \dots \dots . (i i)

a = 5 b = 0

a^{2} + a b + b^{2} = 5^{2} + 5 \times 0 + 0^{2} = 25

I s i t p o s s i b l e ?

Answered by mehdee1342 last updated on 08/Sep/24

(a - b) (a + b) = 25 \Rightarrow a - b = 5

(a - b) (a^{2} + a b + b^{2}) = 625

\Rightarrow a^{2} + a b + b^{2} = 125 ✓

Answered by som(math1967) last updated on 08/Sep/24

a - b = \frac{25}{5} = 5

a^{3} - b^{3} = 625

a^{2} + a b + b^{2} = \frac{625}{5} = 125

(b) ✓

Commented by som(math1967) last updated on 08/Sep/24

a = 5, b = 0 t h e n a^{3} - b^{3} = 125

b u t g i v e n a^{3} - b^{3} = 625 ? ?

Answered by Frix last updated on 08/Sep/24

We know from Linear Algebra :

3 equations

2 unknowns

\Rightarrow

Solve 2 equations and test the 3^{rd}

Or graph it :

(1) f : b = 5 - a

(2) g : b = \pm \sqrt{a^{2} - 25}

(3) h : b = \sqrt[3]{a^{3} - 625}

\Rightarrow no common intersection .

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