Simply expand it and compute the discriminant

Algebra Level 1

(b+ca)(a+bc)(ab+c)(a+b+c)×(abc)(bac)(cab)(abc)\large{\color{#624F41}{(\color{#3D99F6}{b}\color{#20A900}{+c}\color{#D61F06}{-a})(\color{#D61F06}{a}\color{#3D99F6}{+b}\color{#20A900}{-c})(\color{#D61F06}{a}\color{#3D99F6}{-b}\color{#20A900}{+c})(\color{#D61F06}{a}\color{#3D99F6}{+b}\color{#20A900}{+c})\\ \times (\color{#D61F06}{a}\color{#3D99F6}{-b}\color{#20A900}{-c})(\color{#3D99F6}{b}\color{#D61F06}{-a}\color{#20A900}{-c})(\color{#20A900}{c}\color{#D61F06}{-a}\color{#3D99F6}{-b})(\color{#D61F06}{-a}\color{#3D99F6}{-b}\color{#20A900}{-c})}}

Suppose a,b,ca,b,c are integers, is the above product a square?

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